3 Short Walking Breaks Can Reverse Harm From 3 Hours of Sitting 176
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."
+10 health for smokers (Score:5, Funny)
since smokers tend to smoke literally every hour and a cigarette takes 5-6 minutes to smoke.
Re:+10 health for smokers (Score:5, Funny)
+10 for axious pacers
+10 for nervous fidgeters
+10 for furious masturbaters
+10 for immature being my captcha
So smoking is good for you? (Score:3, Interesting)
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So smoking is good for you?
Probably not, no. Don't be stupid.
Re:So smoking is good for you? (Score:5, Funny)
I mean if I get up and go outside for a quick drag once an hour, that's a five minute walk right there.
Not quite sure how the drag helps, but getting changed back into your work clothes afterwards would probably count as exercise.
You should see (Score:3)
You should see the smiles on the faces of slashdotters as they read this news. Seriously, hack into their computers and activate their cameras.
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He meant the camera your wife put behind you.
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Microphone still working?
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Sounds like a yes, then.
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A vague splotch of color moving about isn't much more informative than a black screen.
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You weren't a teenager during the 90's, were you?
Vague splotch of color, not even moving, was easily enough information...
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Re:You should see (Score:5, Funny)
Seriously, hack into their computers and activate their cameras.
Would you please? That would be awesome. I'm running an older version of slackware on this PC, and can't get the camera to work.
Just WATCHed (Score:3, Informative)
What's "Easy" About This? (Score:2)
I don't know too many bosses who would be cool with that.
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Wow, I want to work where you do...where you get up out of your chair and sit immediately back down in some sort of transportation device that takes 5 minutes to get you out of your office so you can walk. ...or you could just get up, walk around the office for 5 minutes, and sit back down, like TFA was saying. =p
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He may be taking into account the time it takes to mentally get back "into the zone". I personally get interrupted enough during a day that a couple of extra times don't really matter (I already go to get water, hit the head, or get lunch, several times a day; it would only take a couple of additional 5 minute walks to get an average of one per hour).
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I already go to get water, hit the head, or get lunch, several times a day
Damn, how many lunch breaks do you need?
He's a hobbit, so he starts with second breakfast, and it just continues from there.
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It depends. If the cafeteria has multiple good dishes, I gotta get 'em all! Fortunately this is somewhat rare :)
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Standing in front of an elevator, standing in the elevator and then a busy escalator, for five minutes down and five minutes back, don't count as walking
Yeah, that would be a real problem if building only had elevators. Buildings are required to have stairs and in most cases they are
publicly accessible. The summary specifically mentions a "leisurely 5 minute walk" so it may be as simple as walking to the
restroom/coffee and back once an hour. If that doesn't take quite long enough then go to the coffee/restroom one floor down.
I know when I worked at HP which wasn't a too big of building just getting from one end of the building to the other took more
than
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I hear you mate, give the office mates a break and leave that growler in another dept.
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Using a farther bathroom is easy. The near one is only a toilet and the arses in the office have nasty shits. Plus the one I go to is a urinal and I despise the sound of piss hitting toilet water. The urinal is in a locker room so you get the sharp smell of BO but it beats the foul odor of shit.
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Why would you want to work somewhere else when you can get paid for just sitting around? If you're worried about the company going out of business, there's not much difference between looking for a new job now and looking for one later.
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What's this "leave the office" and adding 5-10 minutes bit? As soon as you stand up, your 5 minutes starts, and it only ends when you sit down again. Walking down the corridor counts. Walking down and up the stairs if your office isn't on the ground floor counts extra.
Total time required = 5 minutes.
Besides, even shorter periods will help. I believe Apples Watch gamification of fitness targets one minute of standing/walking for each hour of sitting. Which would certainly be an improvement for a lot of offic
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Do you know how far it is possible to walk in 5 minutes? Even if you're not particularly hurrying?
What's "Easy" About This? (Score:2)
Jane/Lonny Eachus goes Sky Dragon Slayer (Score:2)
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
Jane/Lonny Eachus goes Sky Dragon Slayer (Score:2)
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 [archive.today] 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:
Jane/Lonny Eachus goes Sky Dragon Slayer (Score:2)
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:
Jane/Lonny Eachus goes Sky Dragon Slayer (Score:2)
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.
Jane/Lonny Eachus goes Sky Dragon Slayer (Score:2)
Jane/Lonny Eachus goes Sky Dragon Slayer (Score:2)
Jane/Lonny Eachus goes Sky Dragon Slayer (Score:2)
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 radiat
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The REASON there would not be as great a power DIFFERENCE if the chamber walls were also at 150F, is that the walls would themselves be radiating more power out, so there would be less heat transfer (in that case 0).
It is NOT, as you assert, because the heat source would be using less power. That's false, by the S-B equation. Its power output remains the same because (Spencer's stipulation) the power input remains the same.
The reason my solution do
Jane/Lonny Eachus goes Sky Dragon Slayer (Score:2)
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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.
Are you REALLY the moron you make yourself out to be? NET radiation from a cooler surface that passes the boundary is reflected, transmitted, or scattered and passes right back out through the boundary. This is a corollary of the Stefan-Boltzmann radiation law, which states that NET heat transfer is always from hotter to cooler.
You can draw the boundary right around the heat source. Electric power comes in, radiative power goes out. There is no contradiction, and no inconsistency.
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.
And again: by that same
Jane/Lonny Eachus goes Sky Dragon Slayer (Score:2)
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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.
Nonsense. It would take power to bring the chamber walls up to 150F (338.71K). How else do you expect them to get to that temperature? Where are you getting that power from? This is so utterly obvious that I honestly don't believe you don't get it.
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.
You could, but we haven't. Regardless, it still remains the same. Power output at that temperature remains constant because P = (emissivity) * (S-B constant) * T^4 says it has to.
The only thing you are doing is ADDING energy to the system by putting it in an am
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Jane/Lonny Eachus goes Sky Dragon Slayer (Score:2)
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 s
Jane/Lonny Eachus goes Sky Dragon Slayer (Score:2)
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 require
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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?
No. Not right. Since the chamber walls are COOLER than the heat source, radiative power from the chamber walls is not absorbed by the heat source. Because the only power transfer taking place here is heat transfer, which is a function of (emissivity) * (S-B constant) * (Ta^4 - Tb^4).
You DO know what a minus sign is, yes?
Since emissivity doesn't change the input required to heat source to achieve 150F is constant, regardless of where it comes from. But as long as the walls of the chamber are cooler tha
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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.
That is neither correct, or an answer to my question.
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I have already found the solution to a reasonable degree of precision. Your solution, as stated (approximately 241 degrees F for the central heat source) does not check out, even using your own equations.
Jane/Lonny Eachus goes Sky Dragon Slayer (Score:2)
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.
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That was all I needed. I am now done. Have a nice day. You can have the last word all you like; it won't make you any more correct.
Jane/Lonny Eachus goes Sky Dragon Slayer (Score:2)
... 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 [slashdot.org] using
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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.
No, that is not correct. You made assumptions that are, to be blunt, bullshit nonsense.
Since the emissivity for every object in our system is the same, power output is proportional to the T^4. Period. End of story.
Draw your boundary around the heat source. Power in = power out (your own principle). Therefore the power in is 41886.54 Watts, which is the power initially being radiated out.
SPENCER stipulated that this power is held constant. It wasn't my idea. It's a condition of the experiment.
By
Jane/Lonny Eachus goes Sky Dragon Slayer (Score:2)
No. Once again, in this experiment [archive.today] there is a "... constant flow
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No. Once again, in this experiment [archive.today] 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. ..."
I have said nothing that contradicts this. Not only do I freely admit this, my calculations relied on that fact. I kept the power (and hence energy over time) input into the plate from the electric heater completely constant. Which we may freely do, since it was a stipulation of Spencer's experiment.
Jane's even stumbled across this point:
No, I didn't "stumble" over that point, YOU are stumbling over it. Everything changes at thermal equilibrium. The "heated" body is no longer warmer than its surroundings and can begin taking on energy from its s
Jane/Lonny Eachus goes Sky Dragon Slayer (Score:2)
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 requi
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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?
If Ta=Tb, you're doing a different experiment. I've already stated that at that point, it requires no electrical heating power. But it's a straw-man for at least 2 reasons:
[1] it still requires the same amount of power, but once Ta=Tb, it can draw that power from the environment. Before that it can't, because Ta^4 - Tb^4 is a positive number so no net radiant energy is absorbed by (a) from (b). That means all the way up to the exact point thermal equilibrium is achieved, all radiant power is a result of
Jane/Lonny Eachus goes Sky Dragon Slayer (Score:2)
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 -
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The fact that your "global warming" religion will not let you accept the reality of the Stefan-Boltzmann radiation law is not my problem. But you have sure as hell tried hard to make it everyone else's problem.
Ironically, Jane's still insisting that no radiation at all is absorbed by the warmer body.
No NET radiative energy. I did not claim "none at all", and I have repeatedly pointed this out to you. Just no NET transfer from cooler to warmer. This is a fundamental requirement of therm
Jane/Lonny Eachus goes Sky Dragon Slayer (Score:2)
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 omi
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Jane's equation claims "none at all":
electrical power per square meter = (s)*(e)*Ta^4
NOW what kind of bullshit are you trying to pull?
Do you understand what NET means, or do you not? I assure you that a lot of people do. You claimed before that you did.
Why are you doing this? Are you really trying to make yourself look more ridiculous than before?
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.
Repeat: this ASSUMPTION of yours that the chamber walls must be accounted for in the power requirement of the heat source is a direct violation of the Stefan-Boltzmann law. There are no 2 ways around it. Established physics (the Stefan-Boltzm
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I assure you that at least at out current level of technology, we have not managed to build such a sucking device. The heat source radiates out what it radiates out, and nothing around it is "sucking" any power from it.
Although you seem to be doing your very best at "sucking" my time away
Jane/Lonny Eachus goes Sky Dragon Slayer (Score:2)
Again, radiative power out
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power in = electrical heating power + radiative power in from the chamber walls
NONSENSE. The power output is not dependent on the chamber walls, therefore the power input is not dependent on the chamber walls. You're contradicting yourself, trying to have it both ways.
Radiation from the cooler walls has no effect on the heat source whatsoever. This is a basic requirement of thermodynamics!
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:
What's ridiculous is your constant repetition of this bullshit idea. Yes, the cooler walls radiate inward but they have no effect whatsoever on the heat source. ALL of that radiation is reflected
Jane/Lonny Eachus goes Sky Dragon Slayer (Score:2)
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 p
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Ironically, Jane's still insisting that warmer objects absorb no radiation from colder objects. Otherwise Jane wouldn't repeatedly [slashdot.org] object [slashdot.org] to including a term for radiation from the chamber walls in his calculation of required electrical power.
NO!!! This is just plain bullshit. I do NOT object to a term for electrical power. I simply asserted a physical truth: in our isolated system, the electrical power to the heat source, called for by Spencer, has zero dependency on the chamber walls.
It is this nonsense dependency on the chamber walls that I have disputed, nothing else. That is a violation of the Stefan-Boltzmann law.
So just to be clear: I don't object to a term for "electrical power" and never have. My only objection is your insistence
Jane/Lonny Eachus goes Sky Dragon Slayer (Score:2)
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I never said Jane objected to a term for "electrical power". I said Jane repeatedly [slashdot.org] objects [slashdot.org] to including a term for radiation from the chamber walls in his calculation of required electrical power. And Jane continues to do this:
Apparently you did not read what I wrote:
NO!!! This is just plain bullshit. I do NOT object to a term for electrical power. I simply asserted a physical truth: in our isolated system, the electrical power to the heat source, called for by Spencer, has zero dependency on the chamber walls.
What I object to is your insane insistence that the electrical power to the heat source requires a term for the chamber walls. This is sheer nonsense. Standard, textbook physics says the thermodynamic temperature of the heat source, since it is "the hottest thing in the room", as it were, is independent of radiation from the chamber walls. Since it cannot absorb net radiative power from the chamber walls, any electrical power calculation is similarly independent.
Y
What about standing? (Score:3)
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.
Standing Desks? (Score:5, Interesting)
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!
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There were two of them at a furniture store near me. One was well over $1000, the other was well over $2000.
I don't think that most employers are going to spend that kind of money for just a desk. Remember, the inventor of the cubicle originally intended for the furniture to be dynamically changable like that, but cost constraints got it turned into the barely-modular, difficult-to-change setup that we have today.
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I've seen a few employers do this for their workers. LIke 2 out of the 7 different places I've been at (but this is Canada we're talking about). You're right though. Most employers will probably not do this.
The other thing I've heard of for ergonomics is that we should have a chair that lets you lean back and forward spontaneously, rather than have to fiddle with any levers, etc. Supposedly you should relieve the pressure on your belly once in awhile when you sit too...
Anyways, I guess for us office wor
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Eh I did my own with IKEA shelves. Two 8 partitioned KALLAX shelving units with a wide board suspended between them. Found a treadmill on Craigslist for $80, whipped up something to hold my wireless keyboard/trackpad with the handles. It's holding up 2 x 27" screens and I can comfortably walk and type several hours a day.
Not bad for $210.
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Christ life in the US must be bad.
I can't remember the last time, where I saw an office where people did not have those desk.
Also, $1000 is chump change compared to the costs of an employee and even worse a sick employee, so it doesn't make sense to save a few dollars there.
I have one (Score:2)
Then again, I'm self-employed.
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One of my colleagues bought a standing desk from a Kickstarter. So search for that. About $100, all cardboard (and not flimsy - weights 20-30 pounds and he can sit on it without a problem)/
Can't beat that price
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The suggestion was desks that converted between states easily. Around here the best I get is a desk that's not quite tall enough to stand at comfortably and a chair that's too tall to sit in comfortably. Taken together, it's roughly equally comfortable standing or sitting.
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I have a lot more surface area with that $530 desk than I'd get with a $2000 height-adjustable desk though. It'd probably cost a hell of a lot more than $2000 for a desk as big as mine to be height adjustable.
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Maybe go even a step further: treadmill desk.
I had one built and it is awesome.
Learn to sit properly (Score:2)
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 ener
MEH, not impressed. call me back when ... (Score:2)
ah (Score:3, Insightful)
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.
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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.
Hopefully not in the same container. :P
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once to put it back.
I am so not using your coffee machine.
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How about if he said:
Once to get it, and once to get rid of it.
or
Once to get it, and once to make room for more.
Smoke break (Score:2)
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.
Justifies my coffee breaks (Score:5, Interesting)
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Don't know about the rest of you /.tters (Score:2)
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WorkRave.org (Score:2)
just crank it up the music and dance (Score:2)
No need to go anywhere!
Isometrics? (Score:2)
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...
About that (non)-Typo (Score:2)
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.
Re: Do you really think corporations are going to. (Score:2)
Does your boss really micromanage your time that much?
Look for a different job. Yeesh.
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My boss is in a different country with a 9 hour time lag.
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How long do those trips to the bathroom or for coffee take? You might already be getting your hourly 5 in...
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Stop gossiping.