September 12

Here is a link showing the derivation of Mass of Earth

http://www.enchantedlearning.com/subjects/astronomy/planets/earth/Mass.shtml

Please look at the step#1 of the derivation

Since gravitation force "F = GmM/r2" is in between two masses therefore how come this F = GmM/r2 = ma . I mean F = ma is the Weight Force of resting mass acting towards the center of earth and shouldn't be confused with gravity force between masses [F = GmM/r2].

As per separate anaylaysis, shouldn't the weight force of earth be equal to weight force of resting mass. i.e

mo x ge = Me x go Where

mo = Mass of resting object, ge = "g" of Earth, Me = Mass of earth, and go = "g" of resting mass

Even if F = ma is correct then still the Mass of the Earth varies with increase in the size of resting mass on it. This is because of increase in the on-center distance between the two masses e.g moon on the earth. Therefore in this case center to center distance between earth and moon will be

d = Radius of earth + Radius of moon. Thus

Mass of Earth will be = M = [g x d^2]/G instead of M = g x R^2/G, where R = Radius of earth 68.147.41.231 (talk)Eccentric Khattak #1 —Preceding undated comment added 01:34, 12 September 2011 (UTC).[reply]

F=ma is always true in all cases. See Newton's second law. You are using it wrong in this case, since it has nothing to do with gravity. --Jayron32 01:44, 12 September 2011 (UTC)[reply]

The derivation is correct. The "weight force" and the "gravity force" are indeed the same force. I only point out that the equation should read F=mg=m x 9.8m/s² instead of F=ma to avoid confusion with Newton's second law. Your other point is correct. The formula M = [g x d^2]/G works but you must remember to plug earth's gravitational acceleration on the Moon which is much smaller than earth's surface gravitational acceleration. It's much easier to use M = g x R^2/G since in that case the value of g is known to be g = 9.8 m/s². Dauto (talk) 04:00, 12 September 2011 (UTC)[reply]

This means by using F= GMm/d^2, the weight of an object decreases with increase in its size on the surface of earth.

Since moon is resting on earth's surface [ theoritical osculation of two spheres]therefore shouldn't the earth's surface gravity be used in calculation of weight of the moon on the surface of earth as moon acts as a point load on the surface of earth at the common tangent [theoritically].68.147.41.231 (talk) 04:34, 12 September 2011 (UTC)Eccentric Khattak#1[reply]

As I said above, the weight of the moon, unlike its mass, is not an intrinsic property of the moon. It's weight is a consequence of earth gravitational attraction. Given the size of the moon, even if it were on the surface of the earth, its center would be further out and it would be incorrect to simply use earth's surface gravity g=9.8m/s² in the equation F=mg. Dauto (talk) 05:09, 12 September 2011 (UTC)[reply]

Umm.. Sounds like F = mg = GMm/d^2 is lopesided towards earth [only to me].

We know that weight is a measure of the gravitational force exerted on that material in a gravitational field. So,

Is Weight Force = W = F = GMm/d^2 possible without contact of two masses if gravity force is indeed a weight force [You said earlier]?

Example: Would the following gravity force "F" between Earth and Moon be equal to weight "W" of Moon on Earth if a universal law of gravitation is applied?

F = GMm/d^2, Where M = Mass of Earth, m= Mass of Moon, d = Distance between center of Moon to the center of Earth = 356400 km to 406700 km and G = Universal Constant 68.147.41.231 (talk) 20:36, 13 September 2011 (UTC)Eccentric Khattak No.1[reply]

What I said before is correct. I suppose I could also add that weight is NOT a contact force. The contact force is called normal force and in many circunstances (but not always) will be equal in magnitude but opposite in direction to the weight force. Dauto (talk) 21:48, 13 September 2011 (UTC)[reply]

"Thanks Dauto" — Preceding unsigned comment added by 68.147.41.231 (talk) 22:40, 13 September 2011 (UTC)[reply]

a+3b→p enthalpy(-2x kj/mol)/m→2q+r enthalpy(x kj/mol)/ if these reactions are carried out simultaneously in a reactor such that temperature is not changing. if rate of disapbpearance of (b) is y Msec^-1 then rate of formation of q is — Preceding unsigned comment added by Chotu23 (talk • contribs) 04:31, 12 September 2011 (UTC)[reply]

Please do your own homework.

Welcome to Wikipedia. Your question appears to be a homework question. I apologize if this is a misinterpretation, but it is our aim here not to do people's homework for them, but to merely aid them in doing it themselves. Letting someone else do your homework does not help you learn nearly as much as doing it yourself. Please attempt to solve the problem or answer the question yourself first. If you need help with a specific part of your homework, feel free to tell us where you are stuck and ask for help. If you need help grasping the concept of a problem, by all means let us know.--Jayron32 04:48, 12 September 2011 (UTC)[reply]

What would the land area Antarctica be if you could magically disappear the ice? The wiki article lists the ice-free area, but I'm not sure if that means the area that does not have ice on it now. --CGPGrey (talk) 09:20, 12 September 2011 (UTC)[reply]

It depends on the time scale. This blog shows both land above the mean sea level now, and expected land after the isostatic rebound. --Stephan Schulz (talk) 09:26, 12 September 2011 (UTC)[reply]

Interesting. I'd like an estimate of the land area now, if the ice were removed. --CGPGrey (talk) 10:16, 12 September 2011 (UTC)[reply]

That is clearly stated at our Antartica article. —Akrabbim^talk 11:28, 12 September 2011 (UTC)[reply]

Actually, having looked at that article myself, I don't think it is stated very clearly (though I may have missed it): the figure I see is ambiguous as to whether it refers to just ice and ice-free rock, or whether it includes permanent sea ice-cover. Given that the instant disappearance of the ice is not a real-world possibility, however, the effects of a (~60-metre) sea-level rise from melted ice, and the isostatic rebound already referenced by Stephan Schulz, complicate any answer. {The poster formerly known as 87.81.230.195} 90.197.66.53 (talk) 12:11, 12 September 2011 (UTC)[reply]

Ah, sorry. I misinterpreted the numbers given in the infobox. —Akrabbim^talk 17:14, 12 September 2011 (UTC)[reply]

I guess the answer that I'm really looking for is this: under the ice Antarctica is really an archipelago. Is the largest of its islands smaller than Australia? I've done some measurements with Google Earth and the answers looks, pretty clearly, to be 'yes' -- but I'd like to find something a bit more definitive. --CGPGrey (talk) 15:06, 12 September 2011 (UTC)[reply]

We have an article with a beautiful picture. SamuelRiv (talk) 00:50, 15 September 2011 (UTC)[reply]

It used to be expectation that once fission power were developed, it would transform society by providing power that's too cheap to meter. That obviously hasn't happened, but is it still a viable dream? That is, if the safety and proliferation concerns with nuclear reactors could be completely solved, could nuclear plants provide power that's too cheap to meter? --140.180.16.144 (talk) 12:19, 12 September 2011 (UTC)[reply]

Unlikely. A significant part of the cost of electricity is depreciation and maintenance of electrical installations. Even if you posit a "free" source of energy at the power station, the other cost is not going away. One could imagine a "flat rate" model, but that's unlikely, given that a normal household has very strongly fluctuating energy consumption. --Stephan Schulz (talk) 12:39, 12 September 2011 (UTC)[reply]

If it was truly only infrastructural costs, one could imagine that being done based on tax revenue. (In a hypothetical future where tax revenues are still considered a good way to deal with public commons.) --Mr.98 (talk) 12:50, 12 September 2011 (UTC)[reply]

One correction: "Too cheap to meter" wasn't about fission. It was about magnetic confinement fusion. --Mr.98 (talk) 12:48, 12 September 2011 (UTC)[reply]

I don't think we will ever see completely unlimited packages for electricity. Even if the energy source itself was free building the power stations and upgrading the transmission and distribution network to support more power usage will cost money.

I could see how electricity is metered changing though if the economics of production shift. For example in a situation where most power came from plants that were expensive to build but cheap to run you could have a package where electritity in off-peak periods is free or nearly so but peak time electricity is expensive. Plugwash (talk) 13:09, 12 September 2011 (UTC)[reply]

Bananas are free of charge in the jungle, because the factory that produces bananas operates without human labor. So, you can imagine that in the near future we could build a fully automized system that is able to produce electricity and maintain itself.

A solar panel factory that is able to maintain itself and grow all by itself using the electricity provided by the solar panels and the raw materials it can find will eventually cover the entire Earth. Count Iblis (talk) 16:25, 12 September 2011 (UTC)[reply]

That model doesn't really work on Earth, as the land is already spoken for, so you have to buy or rent it. However, in space we could conceivably one day have solar-powered robots that mine asteroids and create more solar-powered robots, passing the excess energy on to us. This is likely centuries away, though. StuRat (talk) 16:33, 12 September 2011 (UTC)[reply]

As long as they don't become sentient and try to enslave or destroy the human race when that happens. Fighting a protracted war against self-sufficient machines would be a real drag on the economy. --Jayron32 18:20, 12 September 2011 (UTC)[reply]

Good idea. Which colour shall we paint it? I think we really need to put some intelligence into the design before we start. God knows what will happen if we just leave it to evolve naturally. Hans Adler 16:35, 12 September 2011 (UTC)[reply]

Note that we still meter things which are really cheap, like water, and even air in some cases (like air for car tires or scuba tanks). StuRat (talk) 16:37, 12 September 2011 (UTC)[reply]

You aren't really paying for air in car tires and scuba tanks, you are paying for pressurized air which is a different animal. Air is abundant, pressurized air, not so much. Googlemeister (talk) 18:18, 12 September 2011 (UTC)[reply]

And you could make the exact same argument for energy. You aren't paying for energy alone, when you buy electricity you are paying for it to be delivered to your home continuously in a form which can be readily used. Energy, is, of course, free, in some forms, such as sunlight. Trees rarely get power bills, as they are known to be rather negligent in paying them. StuRat (talk) 18:23, 12 September 2011 (UTC)[reply]

I too would neglect to pay my power bill if the company insisted on no night access. Googlemeister (talk) 19:50, 12 September 2011 (UTC) [reply]

You can always get interruptable power on your A/C circuit, which means you can use your A/C any time you want, except during power emergencies, which only happen whenever it gets hot. :-) StuRat (talk) 23:09, 12 September 2011 (UTC) [reply]

I've heard utility planners laugh at the Strauss "too cheap to meter" remark from the 1950's and wonder how he could have been so ignorant and stupid. If Strauss had provided a utility with a nuclear generator which provided free unlimited power in 1954, it would still have cost many millions to build the transmission substations and transmission lines to convey it to the load centers, and millions more to construct and to maintain the distribution infrastructure. It all wears out or is damaged by weather, even if underground, and needs expansion and reinforcement, and the workforce and suppliers will not work for free. My electric bill breaks the total cost up into several parts, and the energy cost is only just over a third of the total. "Too cheap to meter" would only work if it did not require infrastructure, administration and maintenance, such as some little gray box on the wall of the house or business, owned by the homeowner or building operator, which provides all the power needed with no fuel or utility connection. Edison (talk) 19:49, 12 September 2011 (UTC)[reply]

Well, if we had gone with Nikola Tesla's crazy scheme to broadcast electricity rather than deliver it by wire, then the difficulty in metering it would have required that we give it away for free, which would really mean it would be paid for with taxes. Note that there is free WIFI in some areas, which is a somewhat similar concept. StuRat (talk) 23:13, 12 September 2011 (UTC)[reply]

In defense of Lewis Strauss (which is an odd thing to worry about, since I find Strauss to be a loathsome character on the whole), "too cheap to meter" is not synonymous with "free". It just means that it isn't worth the effort to monitor usage closely. We have many consumption models in our current society that are not "metered" (most road and highway use, for example) and yet are not at all "free". It should also be remembered that he was speaking in the earliest days of nuclear fusion optimism, when people thought they'd be pulling fuel out of seawater within a decade and things like that, with small-sized plants and so on. It turns out fusion is harder than that and will likely require considerably larger plants at the very minimum. --Mr.98 (talk) 11:52, 13 September 2011 (UTC)[reply]

But electricity is quite easy to meter, at least when delivered by wire. Perhaps road usage is becoming as easy, now, with the ability to electronically identify cars as they drive by at full speed. Myself, I always objected to being constantly stopped, more than to paying the tolls. StuRat (talk) 02:48, 14 September 2011 (UTC)[reply]

Note that if your energy cost is only 1/3 of the bill, this is already quite close to the point of being "too cheap to meter". It is easy to picture that if electricity were ten times cheaper - 1/30 of the bill - the metering would be abandoned. Maybe even if it were five times or three times cheaper. This might be worth doing for marketing reasons alone, but in the days when electric meters had to be read manually, when computing the rate was difficult, when printing and mailing bills was not automatic, it would have made even more sense. Of course, there is also the risk that people would haphazardly throw away electricity without a moment's thought (though the design of appliances that use more power when they're off than when they're on suggests we're not far from that anyway). My guess is that at the boundary line the scheme would have involved a lack of "metering" in the sense that actual usage wasn't measured, but that homes would be subject to the tyranny of a master circuit breaker limiting the maximum current drawn, perhaps more stringently than with metered power. Wnt (talk) 18:27, 14 September 2011 (UTC)[reply]

Note that household wiring already has a master circuit breaker limiting the maximum current drawn, for obvious safety reasons. 67.169.177.176 (talk) 01:12, 15 September 2011 (UTC)[reply]

How often does the Urinary bladder need to be emptied in humans? It does not say the last time I looked in [[Urinary bladder]]. Thank's, and please offer answers even if this particular question has already been answered.213.107.74.132 (talk) 16:30, 12 September 2011 (UTC)[reply]

That's obviously going to vary dramatically based on what fluids you consume and how much moisture you lose in other ways, such as sweat and respiration, which depends on the temperature, humidity, and your clothing and level of activity. StuRat (talk) 16:35, 12 September 2011 (UTC)[reply]

Typical urine output is on the order of 1-2 liters per day, according to urine—though that will vary widely depending on the level of water consumption, physical activity, or consumption of diuretics. While the maximum capacity of the bladder is generally between 500 and 600 mL [1], most people will feel the desire for micturation at about half that capacity. Doing the math, that's three to six times per day, as a rough estimate. TenOfAllTrades(talk) 16:46, 12 September 2011 (UTC)[reply]

You have not answered my question: How often, in hours, does the urinary bladder need to be emptied? Every 2,3,4,5,6,7 hours?. P.S Why does it not say in the article Urinary bladder?--213.107.74.132 (talk) 16:50, 12 September 2011 (UTC)[reply]

If you take 24 hours and divide it by Ten's 3-6 times a day, you get every 4-8 hours. StuRat (talk) 17:04, 12 September 2011 (UTC)[reply]

I'm not counting during the night. Just like 7-11 or so--213.107.74.132 (talk) 17:13, 12 September 2011 (UTC)[reply]

Goodness gracious. Figure it out yourself; you've been given all the information. Do we really have to do something this simple for you?!? --Jayron32 17:16, 12 September 2011 (UTC)[reply]

The bladder only needs emptying when it is full. The amount of times in any one day when it gets full will be different from any other day. StuRat above has given you the reasons for this. The production of urine through the kidneys is not at a constant rate, it varies according to a number of factors, TenOfAllTrades has given you those. It does not give the information you seek in the Urinary bladder article because it is impossible to say given the wide range of circumstances that affect urine output and bladder filling. The urinary system is not a time based mechanism or system, it responds to the level of hydration of any particular person at whatever time of the day, season or year. This is original research - but bear with me. In summer I usually pee about three times from 7am to 6pm, at the weekends I pee a few times more when I have some beers, during the winter when I sweat less I pee about 5 time during the day. Does that clarify the point? Richard Avery (talk) 22:12, 12 September 2011 (UTC)[reply]

  How about this?
  No sweating
  20oC
  5-7 drinks per day

How long would it take for the bladder to fill?--213.107.74.132 (talk) 09:38, 13 September 2011 (UTC)[reply]

You forgot to mention

  relative humidity
  wind speed
  weight of the subject
  subject's body fat %
  subject's clothing choice on the day in question
  subject's hair coverage
  current hydration level
  diet
  amount of sunshine on subject
  bladder size of subject
  kidney efficiency of subject
  etc...

In other words, there are about 7 billion humans on this planet and you will end up with about 7 billion different results.

Googlemeister (talk) 13:34, 13 September 2011 (UTC)[reply]

My answers are:

Average humidity
        wind
        weight
        body fat
        kidney efficiency
Clothing dependent on season
Hydrated
Normal Diet
300 mL bladder

Hope this helps.

Quote

In other words, there are about 7 billion humans on this planet and you will end up with about 7 billion different results.

There are 6.something billion humans, I thought. Has it gone up to 7 already?--213.107.74.132 (talk) 15:11, 13 September 2011 (UTC)[reply]

Human population says 6.96 billion, so close enough. StuRat (talk) 15:14, 13 September 2011 (UTC)[reply]

Not sure if I have average kidney efficiency, or if my bladder is 300 ml, but given those criteria, it took 2 hours 49 min for mine to fill to the point where I am acutely aware of the need to empty it today. Tomorrow it may well be a different length of time. Googlemeister (talk) 16:16, 13 September 2011 (UTC)[reply]

If a car produces 200BHP and 500lb-ft of torque, would it accelerate faster than a car with 400 BHP and 300 torque at a higher rpm? Both cars are of equal weight and shape. Just the engine is different.--213.107.74.132 (talk) 17:12, 12 September 2011 (UTC)[reply]

The RPM affects the BHP and torque, but doesn't figure in to it, here, since the BHP and torque are already known. Also, are we considering the possibility that the wheels will just spin, or do we assuming we maintain traction at all times ? StuRat (talk) 17:21, 12 September 2011 (UTC)[reply]

I know the rpm affects it, i'm not worried about it. If Car A has 200BHP and 500lb-ft torque at X rpm, and car B has 400BHP and 300 torqu at X rpm, which goes 0-60 fastest and which has highet top speed (MPH)? Both has traction. — Preceding unsigned comment added by 213.107.74.132 (talk) 17:24, 12 September 2011 (UTC)[reply]

Because these figures only provide the peak output, it is impossible to answer this question. Torque and HP are dependent on each other. Torque = (HP x RPM)/a coefficient. If you know the torque and the RPM then you know the HP. These figures are just intended to give a rough idea regarding vehicle performance. The reason torque figures are provided in vehicle specifications is to give a sense of the engine's low RPM output. If two vehicles are identical, aside from the engine, you would need a graph of engine out put over the RPM range (in torque or HP) in order determine which would be faster. --Daniel 18:42, 12 September 2011 (UTC)[reply]

Look at these:

3000rpm 360BHP 630 TQ
4000rpm 497BHP 693 TQ
5000rpm 542BHP 569 TQ REDLINE

Versus this higher-reving engine:

4000rpm 436BHP 572 TQ
5000rpm 509BHP 535 TQ
6000rpm 719BHP 629 TQ

These show 2 Engines, one with higher redline than the other. This shows the curve from 3000rpms before the redline. One has higher TQ, and one has higher HP. — Preceding unsigned comment added by 213.107.74.132 (talk) 18:56, 12 September 2011 (UTC)[reply]

That is true, but it doesn't change what I said. Torque and Horsepower are directly related and these figures confirm it. (630 TQ X 3000 RPM)/5252 (coefficient in this case) = 360. Your question can be boiled down to the very simple; which is better for acceleration, peak horsepower or peak torque? The problem is there isn't an answer. In order to determine which vehicle will accelerate faster, you need to look at the entire torque or hp curve, not just the peak. --Daniel 19:53, 12 September 2011 (UTC)[reply]

How about these: (TQ=Torque in lb-ft)

1000rpm  91BHP 477 TQ
2000rpm 229BHP 601 TQ
3000rpm 360BHP 630 TQ
4000rpm 497BHP 693 TQ
5000rpm 542BHP 569 TQ REDLINE
5000rpm is the redline

versus this one:

1000rpm  41BHP 214 TQ
2000rpm 196BHP 514 TQ
3000rpm 319BHP 559 TQ
4000rpm 436BHP 572 TQ
5000rpm 509BHP 535 TQ
6000rpm 719BHP 629 TQ REDLINE
6000rpm is the redline.

With max traction and equal weight and drag, which will do 0-60mph(0-97km/h) fastest and top speed in MPH (162km/h = 100 MPH).

Is it possible for a car to go more MPH than HP? The TQ always seems to peak before the HP for some reason...

How come a veyron, dispite having a 1,001hp engine, can, at 253mph, go only 13mph faster than the McLaren F1, which has 600-odd hp engine?

Well, the weight (4162 lbs for the Bugatti Veyron versus 2513 lbs for the F1) and gearing would also figure in. Also, at those speeds air resistance becomes a major issue, and is nonlinear, such that huge increases in HP are needed for even small increases in speed. StuRat (talk) 17:23, 12 September 2011 (UTC)[reply]

And the SSC Ultimate Aero has 286 more hp, yet is over 20mph faster than Veyron, as opposed to the Veyron having over 350 more hp than F1. Funny. So it's not air resistance, because the Aero gets another 20mph+ with just 286 more hp.--213.107.74.132 (talk) 17:31, 12 September 2011 (UTC)[reply]

It's not JUST air resistance. There are many factors, some of which I listed, and air resistance is a major one. For comparison, look at the speed difference between a car with 10 HP and 296 HP. You'd expect it to be a lot more than 20 MPH faster, because air resistance has much less effect at such low speeds. StuRat (talk) 18:11, 12 September 2011 (UTC)[reply]

It is almost exclusively friction. At those speeds the only way to make significant gains is by decreasing one of the following. Drag coefficient, rolling resistance of the tires or internal drive train friction. The first is by far the easiest as the other two are probably already optimized in a given application. The SSC goes a little faster due to its slightly lower drag. --Daniel 18:22, 12 September 2011 (UTC)[reply]

Agreed. Note that rolling resistance is proportional to the weight of the vehicle, so that's how weight figures in. Drag isn't strictly proportional to weight, but a heavier vehicle is likely larger, with more cross-sectional area and drag. StuRat (talk) 18:55, 12 September 2011 (UTC)[reply]

Which is why an airplane with a 1100 hp engine will do 320 mph without trouble (no rolling resistance and a good drag coefficient). Googlemeister (talk) 19:40, 12 September 2011 (UTC)[reply]

At these speeds rolling resistance is actually effected more by aerodynamic down force then vehicle weight. You've probably heard those stories about Formula 1 cars being able to drive upside down at high speed. --Daniel 19:59, 12 September 2011 (UTC)[reply]

No, I haven't. You mean they go around a half loop and then stay inverted ? StuRat (talk) 23:05, 12 September 2011 (UTC)[reply]

I'm pretty sure that it is purely theoretical, I doubt anyone would actually try given that any problems would be catastrophic. But if they generate more than 1g of down force, which is definitely attainable at high speed, it is at least physically possible. Some vehicles from the Can-Am racing era even used fans to generate low pressure under the vehicle and suck it to the ground, down force is really vital in racing. For an F1 example, see: Brabham BT46--Daniel 23:13, 12 September 2011 (UTC)[reply]

Where is Evel Knievel when we need him ? StuRat (talk) 23:21, 12 September 2011 (UTC)[reply]

An engineer at Williams did once write in to New Scientist magazine, in response to another letter saying that an article on the subject of racing cars must have its calculations wrong. He confirmed that theoretically one could drive a formula one car on the roof of the wind tunnel. He did not say that it had ever been done. Elen of the Roads (talk) 23:26, 12 September 2011 (UTC)[reply]

Question 2: Is BP, HP and PS the same thing?--213.107.74.132 (talk) 17:18, 12 September 2011 (UTC)[reply]

HP is an abbreviation for HorsePower. BP (actually BHP) is an abbreviation for BrakeHorsePower. PS is an abbreviation for PferdeStärke (German for horsepower). All of this and much much more is detailed in the article horsepower. -- kainaw™ 17:26, 12 September 2011 (UTC)[reply]

That article is too complex. Please explain it in plain english.--213.107.74.132 (talk) 17:37, 12 September 2011 (UTC)[reply]

Actually, the article does explain it pretty well in plain english. Kainaw already answered your question. Does anything remain unclear to you? Dauto (talk) 17:45, 12 September 2011 (UTC)[reply]

In the simplest possible terms, PS, BHP, HP and kw are all measures of work power (frequently used to quantify maximum engine output). The have different values (1 BHP doesn't equal 1 HP), but they measure the same thing. --Daniel 18:27, 12 September 2011 (UTC)[reply]

They are measures of power, not work. Dauto (talk) 19:03, 12 September 2011 (UTC)[reply]

Opps, I thought I fixed that, must have gotten an edit conflict and missed it. --Daniel 19:55, 12 September 2011 (UTC)[reply]

What is the longest possible time it can take for the bladder to fill? And the Shortest possible?--213.107.74.132 (talk) 17:39, 12 September 2011 (UTC)[reply]

A deceased person can go hundreds of years without having a bladder fill. If you insert a catheter into a bladder, you can fill it up in a matter of seconds. In case it isn't clear yet: You are asking medical questions that require a diagnosis of the person involved. We will not diagnose anyone here. -- kainaw™ 18:04, 12 September 2011 (UTC)[reply]

I mean by natural means. — Preceding unsigned comment added by 213.107.74.132 (talk) 18:42, 12 September 2011 (UTC)[reply]

It would depend on how big or small it is and how much or little fluid is consumed and how much or how little you're perspiring, AND whether or not you might have some medical condition that's affecting it. If you're concerned it's the latter, get off the internet and go see your doctor. ←Baseball Bugs ^{What's up, Doc?} carrots→ 18:54, 12 September 2011 (UTC)[reply]

For shortest, it can be just a few minutes. I've noticed that when I really have to pee badly, then, after I do, just a few minutes later I have to pee again. I attribute this to urine failing to drain from the kidneys into a full bladder, which then quickly drains into it once emptied. StuRat (talk) 18:53, 12 September 2011 (UTC)[reply]

For longest, I'd go with indefinitely. Does somebody who is seriously dehydrated still urinate ? (There are also those with kidney failure, some of whom don't urinate at all.) StuRat (talk) 18:53, 12 September 2011 (UTC)[reply]

Urine failing to drain from the kidneys into a full bladder? Isn't this going to damage the kidneys? Count Iblis (talk) 20:38, 12 September 2011 (UTC)[reply]

Yes - see Vesicoureteral reflux. --TammyMoet (talk) 20:56, 12 September 2011 (UTC)[reply]

That's not just failing to drain into the bladder, but moving backwards, from the bladder into the kidneys. StuRat (talk) 23:01, 12 September 2011 (UTC)[reply]

• To reassure StuRat, the reason his bladder refills rapidly in that situation is that the body stores excess fluid in the bloodstream, not the bladder. If it's not convenient to micturate and the bladder fills, the body can keep fluid circulating in the bloodstream. The minute the bladder empties, it ships some of the excess fluid out thru the kidneys and fills the bladder up again. This effect is most noticeable if one has sunk a few pints (of anything), and then waited some time before pointing percy at the porcelain. Elen of the Roads (talk) 23:22, 12 September 2011 (UTC)[reply]

Sometimes known as "cracking the seal". You can go for several pints without visiting the dunny, but once you've cracked the seal you're going every 10 minutes. Any idea of the mechanism of that, Elen? Does a full bladder stimulate ADH production, perhaps? Tonywalton ^Talk 23:32, 12 September 2011 (UTC)[reply]

It takes some time for the liquid to be absorbed into the blood stream, removed by the kidneys, and collect in the bladder. So, once the bladder fills, this process is well underway and will continue until the stomach, blood stream, kidneys, and bladder have all been completely emptied. StuRat (talk) 02:53, 13 September 2011 (UTC)[reply]

And what about the longest time by natural means while alive?--213.107.74.132 (talk) 08:46, 13 September 2011 (UTC)[reply]

Are you seeking medical advice? ←Baseball Bugs ^{What's up, Doc?} carrots→ 09:54, 13 September 2011 (UTC)[reply]

NO!!!--213.107.74.132 (talk) 09:59, 13 September 2011 (UTC)[reply]

According to the History channel, traditional japanese swords such as katana, are not just the just the best swords of their day, but that the ancient swords are even harder and more durable than any sword that can be made by modern techniques.

This seems implausible to me. I'm sure that the ancient swords and their folded composite construction was really impressive in their day (and is still rather impressive today). But I would assume that modern metal workers, with modern alloys and forging techniques, could ultimately do better by virtually any objective set of standards one could choose. Are there objective comparisons between top of the line modern swords and their traditional counterparts? Dragons flight (talk) 22:46, 12 September 2011 (UTC)[reply]

There are two distinct questions here:

1) Could modern science make a better sword ? I assume that they could, with enough research and effort.

2) Do we currently make any better swords ? Here I wouldn't be surprised if the answer is no. This is because modern swords are just for decorative purposes, for the most part, so looking shiny is more important than being useful weapons. Therefore, this gives them very little incentive to put in the time and money such a project would take. StuRat (talk) 22:58, 12 September 2011 (UTC)[reply]

The answer is no. My father was the manager at the Wilkinson Sword sword works in central London for some years in the 1970s, and he was always of the opinion that modern swords were not of the quality of older swords. The reasons seemed to be numerous. The starting product - pig iron - was different. I expect the modern stuff is much purer. The method of steelmaking was different - you're not making it by folding over charcoal, the carbon is injected into the converter (I once went round the Silver Fox plant in Sheffield, that specialised in making this kind of stuff). The steel is rolled into a sheet and blanks stamped out with a press, rather than each sword being formed from a bar of iron. The shaping process is different - uses a rider hammer instead of a blacksmith's arm. The steel no longer undergoes the multiple foldings and quenchings. And the last quenching always used to be in some kind of animal fat (Wilkinsons used whale oil apparently), whereas now it's mineral oil.

That said, you can put a pretty impressive edge on a cavalry sabre made by modern methods. My father posesses a sword called a hankerchief cutter, made for a demonstration at the Edinburgh Military Tattoo I believe in the time of king Edward VII. A cavalry officer would ride around the arena, and apples and oranges would bethrown up to him to cut in half. The climax was a pass in front of the royal box, when a weighted handkerchief would be thrown up and cut in half. Elen of the Roads (talk) 23:15, 12 September 2011 (UTC)[reply]

Modern forging and tempering techniques are indeed better than ancient ones, and could theoretically be used to make swords that are as good as (or better than) ancient swords. However, the crucial difference that makes ALL modern swords inferior to the best ancient swords is the steel. Many ancient swords were made of bulat steel, which can no longer be produced because the recipe has been lost. And when you put a sword made from bulat steel against one made from the best steel made these days, it will be the one made from bulat steel that gives better strength, hardness and cutting power. 67.169.177.176 (talk) 00:06, 13 September 2011 (UTC)[reply]

Correction: several modern metallurgists have been reported to have rediscovered the recipe and manufacturing process for bulat steel; however, their processes at this time are highly experimental and are not used in the quantity production of swords. 67.169.177.176 (talk) 00:22, 13 September 2011 (UTC)[reply]

I personally have tried to look for any modern swords and as far as I could tell, there are none. The closest would be the Shin guntō which is basically a machine made katana, and as such it's inferior to authentic katanas. Can they make a better sword? My question is, how do you improve upon something that is already at the pinnacle of its development? They already perfected the art. There are some science fiction type ideas on how to improve a sword like adding in some kind of vibrating technology, producing a vibrosword that can cut through almost anything, but that only exists in science fiction currently.

Add a cutting chain. :-D 67.169.177.176 (talk) 02:32, 13 September 2011 (UTC)[reply]

How much would you be willing to pay ? I imagine there are craftsman who could make one the old-fashioned way, but they would want something like $100,000 for such a sword. If you're only willing to pay $1000, then you can only afford a machine-made sword. StuRat (talk) 02:48, 13 September 2011 (UTC)[reply]

Is is possible that older swords were made from low-background steel that is no longer available, and that this makes a difference? Mitch Ames (talk) 12:38, 13 September 2011 (UTC)[reply]

Not appreciably. The chemical and physical properties are essentially the same as far as swordmaking is concerned. The amount of additional radioisotopes is far too small to make a difference.

What's missing is experience in the artisanship of making and working the steel. A great part of swordmaking was numerous serendipitous discoveries (dumb luck) accumulated over centuries, as well as practical experience handed down from master to pupil. Both of these have largely been lost because they hadn't been written down or properly understood from a scientific point of view.

If there were ever a need for high-quality weapons-grade swords again, a period of trial and error as well as training would be required. Our current knowledge of metals and metallurgy, and the availability of advanced production and diagnostic technology, would mean that that period would last on the order of years or decades, not centuries, because dumb luck would not be as dumb as it once was, being guided by modern science and technology.

The question is whether we want to faithfully recreate old swords using the processes once used to produce them, or to create swords that are as good or better than them by any means at our disposal. The former would take considerably longer to implement than the latter. A period of training for those who are to use the swords would also be necessary. In the hands of a neophyte (like me, for instance), a superb katana is not appreciably better than a cheap mass-produced replica. Dominus Vobisdu (talk) 13:08, 13 September 2011 (UTC)[reply]

Yes, the arts were lost over time, as swords became less weapons and more decoration, so a cheaper, faster, mechanised process was adequate to give a product that would take sufficient of an edge to allow it to be sharpened and used for what were largely display purposes. The technical drive in steelmaking went into the production of products for shaving. Elen of the Roads (talk) 14:52, 13 September 2011 (UTC)[reply]

FYI, there is still swordmaking in the traditional style in Japan. Iaido sensei often practice with one - one was priced at $8000. SamuelRiv (talk) 00:40, 15 September 2011 (UTC)[reply]

I thought tundra have year round permafrost. My textbook say at summer arctic tundra only upper soil layer melts permafrost. Bottom soil suppose to stay greyish-white and ice suppose to stay. Then if bottom soil stays permafrost then how can those lands get 60F or higher. Actually most tundras get 60F or higher in summer time for like about 5 to 8 consecutive days and someyears top 70F, then how can permafrost not melt and white-gray ice stay at the bottom. Is Kangerlassuaq, Greenland subarctic Dfc Koppen, Trewartha EXld or is it tundra ET or FT. Most geography map said Kangerlassuaq, Greenland is tundra, nowhere said it is subarctic. But the article said the summer average high both July and August nearly top 60F or higher nearly everyday mostly mid 60s F. My understanding is is tundra cannot month average above 50F. July isotherm for Kangerlassuaq, daytime is 61F, low in early 40s. That makes the moth average nearly 51F. That is too high for tundra.--69.229.6.251 (talk) 23:09, 12 September 2011 (UTC)[reply]

Do you have any idea how low the thermal conductivity of soil is? It's perfectly possible for soil to be at 70F on the surface and below freezing several meters down. 67.169.177.176 (talk) 00:26, 13 September 2011 (UTC)[reply]

Agreed. Since I suspect that English isn't the native language of the OP, let me simplify: "Dirt is a very good insulator and keeps the heat above from getting down to the permafrost all summer long, so it stays cold down there." StuRat (talk) 02:41, 13 September 2011 (UTC)[reply]

As far as the definition of tundra, MY understanding is that it's based on vegetation, the presence of permafrost, and (possibly) soil type, not on temperature. 67.169.177.176 (talk) 00:30, 13 September 2011 (UTC)[reply]