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watt's up with that? (actual electric draw less than advertised)edit
OK... so if a horsepower is like 750 watts, and US AC voltage is like 120 volts, and the average outlet is fused at 15 amps, that's 1800 watts, which is 2.4 HP. So, why do so many appliances, many from reliable manufacturers, claim like 4, 5, 6 HP or more from motors which the plate says draw 15 amps or fewer? thanks.
US A/C voltage for resinces is 240V with a center tapped delta transformer. Commercial buildings are often 480 volt Y tapped transformers with 277V volts between pairs. The specific appliance that you are referrimg to is paramount. It is relatively to scale voltages to meet current demands as THHN can handle 600 volts across three phases. --DHeyward (talk) 05:09, 10 February 2014 (UTC)[reply]
Lots of the residential US is wye not delta...see distribution transformer#Connections. It also might be "technically correct but un-necessarily confusing to the lay reader" to talk about average US outlets and then say "240V". Sure, supply is 240V across both ends of the split phase, but only specially wired heavy-load appliance outlets get that (air conditioners and things that have electrical heating). Most normal outlets have a neutral (to the grounded center tap) rather than two hots, and are therefore 120V. Poster mentioned common appliances on 15A outlets; NEMA 6–15 (or L6–15) are not a common household outlet or appliance plug from what I've seen. DMacks (talk) 09:45, 10 February 2014 (UTC)[reply]
Incorrect. A Y split woulb be 208V between hot phases and 120V to the center Y tap. A delta tap is 120/240. Every residential appliance such as A/C or water heater is 240V. The load in a neighborhood is distributed through different center tapped delta transformers (the grounded tap rotates). In commercial applications, the standard is 120/208 (or 480/277) which is a wye configuration. No normal residential structures are designed for 120/208. Instead they are designed for center tapped, single phase 240/120 delta configurations. Load balancing often allows for a reduced neutral conductor size. The delta transformer is 240V across each of 3 phases and only a single, center tapped phase is sent to a house. Ever transformer can center tap any of the phase legs. In the end, the neutral return has relatively zero current. Delta configurations are the only way 120/240V loads centers work. Wye configs are 120/208V and non-existent in U.S. houdseholds. --DHeyward (talk) 10:35, 10 February 2014 (UTC)[reply]
It may be delta at the mid-level step down stages, but the last one before the residential property lines often appears to connect to exactly one of the three "top of the pole" (primary distribution) phases: only one has a wire coming off of it, that goes to one cut-out, that connects to one bushing on top of a pole-mount transformer, that becomes the two hots to a house. That's phase-to-ground, not phase-to-phase...at most, it could be some asymmetric/4-wire delta? The other common arrangement I see is a set of three transformers each similarly wired to one of the three phases, and with a common wire among all three that goes to ground rather than having them attached pairwise. Again, it's exactly as our article says. File:Polemount-singlephase-closeup.jpg is in Canada, but it's the same wiring configuration I see all over the US. The two hots to a property aren't off two phases (that type of wye loading would indeed give nonstandard voltage combinations) but are the two parts of a single-phase split-phase on a transformer whose center-tap is grounded. DMacks (talk) 11:33, 10 February 2014 (UTC)[reply]
I think I see what you're saying. The final transformer is from a single leg primary with a secondary center tap. That is correct. --DHeyward (talk) 15:39, 10 February 2014 (UTC)[reply]
An electric motor's rated HP is the maximum mechanical power output the motor can produce continuously under optimal circumstances. Unfortunately, for some kinds of appliances, such as wet/dry vacuums and air compressors, it's become common to describe the motor in terms of "peak HP". "Peak HP" is the electrical power that the motor consumes during the first few milliseconds after it's turned on, which is much higher than the power the motor normally consumes or produces due to inrush current. Circuit breakers are designed to handle a very brief inrush current like that that's higher than the nominal rating of the breaker. Peak HP isn't a particularly useful measure of how powerful a motor is, except that it's useful for marketing purposes because it's a much higher number than the motor's rated HP, so it sounds better in comparison.[1]Red Act (talk) 08:10, 10 February 2014 (UTC)[reply]
If that's the case then it would still trigger a fast circuit breaker during the few milliseconds of peak power (though I appreciate that circuit breakers may be designed to delay triggering, thereby reducing their contribution to safety). Are there really appliances rated 6 horsepower that have a 120v input? It sounds unlikely to me, but I'm not familiar with US power outlets. Are all the 240v ones permanently wired, or are there outlets that can power up to 4 horsepower continuous (240v 13 amp) as in the UK? On this side of the pond, horsepeower rating are for continuous usage. Dbfirs 12:06, 10 February 2014 (UTC)[reply]
In the US it is common for permanently installed through-the-wall air conditioners (which look like window units but are a bit bigger) to be powered through a dedicated socket. Likewise for electric ranges and electric clothes dryers. All of these should have dedicated sockets which supply 240 V. Electric water heaters and some electric ranges are wired without the use of a plug & socket. The maximum rated current ranges from 15 A to 40 A depending on which kind of appliance the socket is intended for, and each maximum current has a socket of a different shape to prevent plugging a high-current plug into a low-current socket.
All portable appliances I've ever seen in the US, such as vacuum cleaners, fans, toasters, and food mixers use 120 V. Jc3s5h (talk) 16:30, 10 February 2014 (UTC)[reply]
Here's an example of a wet/dry vacuum with the ridiculous labeling. As per the specifications, it plugs into a 120V outlet. The product title at least calls it "6.5 peak HP", but the detailed specification just says "6.5" for "horsepower", without even using the word "peak". There's no indication on that web page as to what the wattage rating or real HP rating of the motor is; all the consumer has to go on on that page is the ridiculous "6.5 peak HP". Red Act (talk) 16:20, 10 February 2014 (UTC)[reply]
I asked more or less this exact question of an actual manufacturer (a vacuum cleaner - about 5 years ago) - they said that just as CFL lights are often described as "60W" when they actually only consume 15W but generate as much light as an old-fashioned 60W incandescent - their motors (whilst only consuming 2.4hp of electrical power) generate as much OUTPUT power as an old-fashioned 4hp motor would have. Sounds like a load of B.S to me...what is the benchmark for an "old-fashioned motor"? But I do see where they're coming from. When devices such as light bulbs and electric motors are described to the consumer by the amount of energy they consume - and those customers think "More Is Better", it's hard to convey that your motor (or whatever) produces more power than the opposition. What you'd really want is to know the input hp AND the output hp - buying a vacuum cleaner that has the smallest first number for the largest second number...alas, they seem to believe that people are too stupid to understand that - so "more is better". Meh. SteveBaker (talk) 16:13, 10 February 2014 (UTC)[reply]
T've seen this in advertising as well as the vacuum cleaner describes itself as equivalent "suckage" amp rating. I've also seen compressors that need a bit of IR drop between the compressor and breaker to prevent it from tripping. It worked as long as it was connected to an extension cord. --DHeyward (talk) 06:59, 11 February 2014 (UTC)[reply]
A helical integrated CFL, one of the most popular designs in North America, since 1995, when a Chinese firm, Shanghai Xiangshan, marketed the first successful design.[1]
I have a helical CFL installed on a cheap IKEA goose-necked desk lamp that looks just like this one. I put it in my living room and the last time I turned it on was last summer. I only use it when I need to replace other more efficient fluorescent or LED light tubes.
The room temperature is about 20C or 70F. In fact the room temperature seldom drops below 15C or 60F. I probably pointed the bulb head up or sideways to light up the ceiling so I can work on the light fixtures. It was placed there unplugged, unmoved and untouched for several months.
I pointed it downward and turned it on today. For the first few minutes I did not take a look at it. Then I noticed something strange.
The head of the bulb remains dark (the top part of the picture). The feet (the bottom part of the picture) is bright. The middle section of it is not very bright.
I turned it off to get my camera. However, when I returned and switched it on again, this helical CFL returned normal. There's no need to take a picture of it.
What may cause this? Can mercury condensation cause it? -- Toytoy (talk) 05:02, 10 February 2014 (UTC)[reply]
I notice a less extreme example of this on all the bright CFL bulbs in my house, they start off only bright at the base and take a good 20-30 seconds before the whole bulb gets to its full brightness. Your example sounds like just a more extreme case of the same thing. Not sure what causes it, I'm guessing it has an operating temperature that it has to reach before the brightness is optimal. Vespine (talk) 06:08, 10 February 2014 (UTC)[reply]
Agreed. Also note that there are "instant" CFL light bulbs that get closer to full brightness quicker. I kind of like the slowly brightening type, though, as it's not such a blinding shock when I turn them on in the morning. See Compact fluorescent lamp#Starting time. StuRat (talk) 07:06, 10 February 2014 (UTC)[reply]
Why do wheels have to be stronger when using disk brakes vs rim brakes?edit
A lot of sites[2][3] say that wheels have to be stronger when using disk brakes vs rim brakes. is this right? Naively it seems that both will be transferring the friction force between tyre and the road to the mass of the moving vehicle, so wouldn't it be the same? -- Q Chris (talk) 09:15, 10 February 2014 (UTC)[reply]
No, it would not. The strength is the ability to cope with torque between the rim and hub not the torque between road and axel. --BozMotalk 11:17, 10 February 2014 (UTC)[reply]
How is this different? The breaking force must come from the road and (ignoring the weight and angular momentum of the wheel) apply on the axel. -- Q Chris (talk) 12:25, 10 February 2014 (UTC)[reply]
If the disc is closer to the hub than the rim, then there will be a higher force there for a given amount of braking. Can disc brakes produce a greater amount of braking than rim brakes? If so, then this also would require stronger wheels (assuming sufficient friction with the road). Dbfirs 12:11, 10 February 2014 (UTC)[reply]
Disk and rim brakes can both usually lock the wheels, its resistance to fading and clearing of water that's better -- Q Chris (talk) 12:25, 10 February 2014 (UTC)[reply]
The concept of "locking the wheels" depends on the friction with the road surface, that's why I said "assuming sufficient friction". I agree that it's usually easy to lock the wheels with efficient rim brakes. If friction is low then forces at the brake will be low for both types of brake. It's still true that, for a given torque, a force nearer to the hub will be greater than a force at the rim. I see what you are asking, though, because if the disc is mounted on the hub, then the spokes will not need to be stronger unless the braking force is higher, and the limit to braking force on the bike (and rider) is usually limiting friction. Dbfirs 12:56, 10 February 2014 (UTC)[reply]
The bulk of the force that needs to be stopped comes from the momentum of the rider. In the rim brake case, the braking force is transferred from the road to the tire to the rim to the brake to the frame to the rider. In the disk brake case, it's transferred from the road to the tire to the rim to the wheel (through the spokes to the hub) to the brake to the frame to the rider. In the disk brake case, since all the force passes directly through the wheel, yes, it needs to be stronger. —Steve Summit (talk) 13:02, 10 February 2014 (UTC)[reply]
I don't see a difference, I am beginging to think that like Dbfirs says "...the spokes will not need to be stronger unless the braking force is higher". -- Q Chris (talk) 13:52, 10 February 2014 (UTC)[reply]
Another way of looking at it is that in the case of rim brakes, much of the braking force is not applied on the axle. (Because, as I said, it is transferred directly from rim to brake to frame). —Steve Summit (talk) 14:51, 10 February 2014 (UTC)[reply]
The only connection between the rim and the frame is through the wheel and the hub. I'm going to post a link to this onto the maths desk and see what comes from that. -- Q Chris (talk) 14:56, 10 February 2014 (UTC)[reply]
Not when you're actively braking with rim brakes. Then there is a connection through the brakes from the rim to the frame. That's what Steve Summit has been trying to explain. Katie R (talk) 15:15, 10 February 2014 (UTC)[reply]
For an intuitive way of picturing it, imagine a bike with completely locked perfect brakes. Something like a bolt through the rim attaching it to a bracket on the frame for rim brakes, or a welded hub for disc brakes. If you were to try to rotate the wheel by applying force to the tire, it certainly seems like the spokes or hubs would fail first on the disc brake system, where the rim itself would have to be torn apart with rim brakes. Katie R (talk) 15:20, 10 February 2014 (UTC)[reply]
I see - it is going through the wheel but to two points - the top and the centre instead of just the centre! -- Q Chris (talk) 15:23, 10 February 2014 (UTC)[reply]
See moment of inertia. Consider a simple illustration: if you are trying to hold a door closed while someone is trying to push it open, do you put your hands near the knob, or near the hinges? This is very similar to disk vs. rim brakes. Pushing near the hinges puts a larger force on the door, and requires more force (compared to pushing near the knob) to keep the door closed. This is why wheels have to be stronger for disk brakes. SemanticMantis (talk) 17:54, 10 February 2014 (UTC)[reply]
You mean moment of force (not moment of inertia which is insignificant when compared with the linear momentum of the rider). This argument applies only to the brake force being transferred to the tyres for friction to slow the bike down. The reduction of the linear momentum of the rider has to be transferred from the tyres, via the hub and the frame, and is the same however the braking is achieved. I see Katie R's argument about an extra force on the brake lever at the rim, but isn't this a forward force? Dbfirs 08:06, 11 February 2014 (UTC)[reply]
Excellent point - unless the wheel is locked the force applied by the brake to the frame is forward force -- Q Chris (talk) 09:56, 11 February 2014 (UTC)[reply]
Yes, I was using the principle that friction always opposes relative motion. Once the wheel is locked I think it will change to a backward force. Dbfirs 11:57, 11 February 2014 (UTC)[reply]
Actually it will still be forward after the brakes lock - think of which way it will move if there is some slippage -- Q Chris (talk) 15:11, 11 February 2014 (UTC)[reply]
Yes, of course it will! I'll re-think my analysis below. Dbfirs 22:55, 13 February 2014 (UTC)[reply]
Oops, yes, I meant moment of force, thanks. And you're right that the reduction of the linear momentum is the same for both types of brakes. But the tangential (i.e. perpendicular to radial) forces acting on the spokes are higher in the disk brakes, requiring stronger wheels. This perspective is in line with Ulflund's comment below. It is relevant because, even though it is linear momentum that is ultimately reduced, this is accomplished by a negative (i.e. opposing the direction of current rotation) torque on the wheel. SemanticMantis (talk) 15:59, 11 February 2014 (UTC)[reply]
I think the explanations above are mostly valid but only explaines small differences. The main difference is that with a disk brake the spokes needs to transfer a torque from the hub to the rim, while with a rim brake they only need to transfer a direct force. Transfering the torque will require much larger forces in the spokes since the hub is small compared to the rim and thus provides only a short lever arm. Ulflund (talk) 09:56, 11 February 2014 (UTC)[reply]
But the rim is not where the rider is connected to the bike, so I don't see how this makes sense. Why would one consider rotation about the rim (unless the rider is going over the handlebars, and then it is a different point on the rim that is stationary). Dbfirs 11:57, 11 February 2014 (UTC)[reply]
I've re-thought my analysis and decided that it’s not sufficient to consider torque and forces on the wheel as a whole. The magnitude of the net torque for both types of brake is almost negligible compared with the large forces of deceleration. I claim this because the angular momentum of the wheel is small compared with the linear momentum of the bike plus rider, so the torque required to reduce this to zero is comparatively small. Instead, if one considers torques from rim to hub, there are two almost equal forces at (almost) opposite points on the rim for rim brakes, and only one for disc brakes, but the magnitude of the torque from the tyre friction with the road will be identical for equal braking. One might imagine that the additional torque from the rim brake would be more likely to distort the wheel, but it acts in the opposite rotational direction across the opposite half of the wheel, and I think that most of the stress will be distributed round the rim, rather than between rim and hub. With the same force just at the bottom of the wheel for disc brakes, the balancing force on the disc acts close to the hub and does not appear across the spokes if the disc is attached to the hub, so a torque is set up between the rim and the hub that needs strong spokes to resist. Also, if the disc is mounted on one side of the wheel, then the fork on that side will need to be much stronger than for rim brakes, and there will be a twisting effect on the wheel about a different axis, which might also explain why the wheel has to be stronger for disc brakes. If two discs are used (one each side of the wheel) then this twisting effect should vanish, but the cost will increase. I've been looking on the web for a proper analysis of braking forces and torques, but I haven't found one. In view of the fact that my previous analysis was faulty, I would be glad if someone could check this analysis. Someone must have either thought this through or done some experiments. Dbfirs 17:18, 14 February 2014 (UTC)[reply]
Woo-ee! Has anyone thought to consider a limiting case? Think of a spoked wheel with radial spokes to a hub of extremely small radius. A torque on the hub resisted at the rim will create distortion, a bit like a fishing reel trying to reel in a lot of lines, and would easily destroy the wheel. This construction will be fine with rim brakes, however, as throughout the braking manoeuvre, the hub is allowed to rotate freely so that it transmits no torque to the rim. Also, assuming a moderately rigid hub, one sided braking on a disc is transmitted essentially distortion-free to the other side; the only component that is likely to have to be designed to handle the one-sided disc braking is the fork. The wheel (other than for the torque felt by the hub tube – and the forces on the axle) will not be affected in any way. —Quondum 01:59, 18 February 2014 (UTC)[reply]
The extent and duration of the present flooding in the UK Somerset Levels caused me to wonder: How long can worms and insects and other small animals survive when their habitat is extensively surface flooded. And what kills them off - drowning, suffocation, starvation, crushing, ...? -- SGBailey (talk) 12:51, 10 February 2014 (UTC)[reply]
... or being eaten by birds? -- lightly flooded fields attract rooks and crows because they know that the worms will come to the surface. Dbfirs 12:58, 10 February 2014 (UTC)[reply]
I see large numbers of drowned worms after heavy rains, but don't see any drowned small animals. From this I conclude that worm's only survival strategy is to crawl to the surface. If the surface is also flooded, they don't climb trees or anything to escape, they just drown. Also, breathing through their skin as they do isn't very efficient, such that just poking their head above the water won't save them.
Ants seem to be able to climb plants to escape. An interesting hijacking of this ability is the fungus which infects them, triggers that anti-drowning response, then kills them once they clamp onto a leaf on a plant, explodes out of their head, and rains down spores on the next batch of ant hosts.
Small animals, on the other hand, presumably find their way to higher ground somehow, and can survive as long as they aren't totally submerged (unless it's cold enough for hypothermia to occur). StuRat (talk) 15:36, 10 February 2014 (UTC)[reply]
For the zombie fungus, I thought the point of climbing was to be eaten by a mammal for an alternate phase of its cycle. —Tamfang (talk) 02:42, 12 February 2014 (UTC)[reply]
In the UK, I think many insects are currently overwintering as eggs or pupae, i.e. forms which are relatively resistant to drowning. In the summer months, die-offs due to flooding, drought, etc. are pretty common. Recall that most animals smaller than your hand are relatively r-selected, and these creatures' populations are generally well-equipped to "bounce back" after large mortality events. (Note, I wouldn't put too much weight on the "outdated" description of r-k selection in our article. It is true that current research uses more nuances frameworks of life history, but the essence of r-k selections is still a generally viable heuristic description -- the essence is that small animals generally have very high fecundity and short generation time. Survival of the individuals doesn't matter much, if there is a giant pool of new of recruits ready to go). SemanticMantis (talk) 16:47, 10 February 2014 (UTC)[reply]
Worms can gain oxygen from the water in the soil as long as it has enough present it will not drown. Chromagnum.
Based on the number of drowned worms I see after heavy rains, I must conclude that this system doesn't work very well for them. Specifically, both lungs and gills take measures to increase the surface area dramatically, to increase the exchange rate. In worms and insects, this doesn't happen, so they need a high oxygen environment. A reduction in oxygen content in the air since the age of dinosaurs is likely also responsible for the reduction in size of such creatures since then. (Alternatively, larger ones can have a very slow metabolic rate, as that also requires less oxygen.) StuRat (talk) 18:17, 12 February 2014 (UTC)[reply]
No distance in the galactic coordinate system?edit
Do I understand well the galactic coordinate system from our article? It seems it has only 2 variables (ℓ and b) and therefore only tells you where to look in the sky, it does not tell you anything about how far the object is? So it would be insufficient as a means of knowing in which direction to go, for some galactic travellers that do not start from the sun? --Lgriot (talk) 14:32, 10 February 2014 (UTC)[reply]
When that system was first established (in the 1950's) it was really only a slight reformulation of the 1932 system. Back then, it was extremely difficult to determine distances to anything at galactic scales - so including a meaningful distance component would have been almost impossible. Also, you have to understand why the coordinate system is needed - it's purpose. You need a way for one astronomer to tell another astronomer: "Go look <here> for this amazing thing I found!" - and for that, you only need a direction to look in. Before the digital age, you'd need to print star catalogs showing where to look for something - and including a distance would be both pointless - and likely to be revised with later measurements. SteveBaker (talk) 16:15, 11 February 2014 (UTC)[reply]
In many places those with solar panels, windmills, etc., can sell excess electricity back to the power company using the normal power lines. I'm curious about how this works. That is, I can't see how it easily can be sent back as A/C matching the wave form of the original electricity, so is it sent back as D/C, instead ? If so, how does this affect the neighbors ? Do their incandescent bulbs burn brighter ? Do devices like clocks which depend on the 60 Hz rate get thrown off ? Do we have an article on this ? StuRat (talk) 16:20, 10 February 2014 (UTC)[reply]
See Grid-tie inverter. They use a computer to match the phase with the power grid. Mr.Z-man 18:03, 10 February 2014 (UTC)[reply]
My next door neighbor got solar installed recently so I was going to point to the inverter page too. Of course even if the power wasn't fed back into the grid, solar panels don't just pump DC straight into your domestic circuit, never mind the neighbors, what would it do to your own appliances? Vespine (talk) 01:14, 11 February 2014 (UTC)[reply]
If you connect a generator to the grid, the direction of power flow to or from the grid depend on the phase angle. If your (mechanical) generator is in synch with the grid, then you put more power to it so that it wants to turn a bit faster, you create a leading phase angle and you push power to the grid. If, on the other hand, you increase the excitation voltage a bit, rather than advancing the phase angle, you would push "VARs" or volt amps reactive onto the grid, providing voltage support to the grid.As for the neighbors synchronous clocks, you would have to have such a big generator that you increased the actual grid frequency to make the clock run faster. As for solar panels, any power would be converted to AC before being pushed into the grid, and again the electronic circuit would have to provide a sinusoidal voltage at a leading phase angle. Such inverters are available in small hobbyist size and I've seen Youtube videos of them in action. If you put 1 kilowatt into the grid, then the flow from the substation would back down by one killowatt (out of perhaps megawatts) and to the utility it would just look like the demand for power decreased by a tiny amount. Utilities have big rulebooks covering distributed generation or cogeneration, usually including requirements that if your generator is so big it might be able to keep the line powered when the substation breaker tripped, then you would have to add utility grade relays to trip your generation when the utility tripped, perhaps including a communication channel to/from the utility for automatic high speed tripping, adding perhaps a hundred thousand dollars to the cost.Edison (talk) 16:20, 11 February 2014 (UTC)[reply]
Thanks all. So they do match the frequency of the A/C signal already there. Interesting. StuRat (talk) 02:57, 13 February 2014 (UTC)[reply]
It can be simpler than that. Any standard synchronous or induction motor for example, will automatically do the job for you, no electronics required. Just apply forward torque to it while it's attached to the supply, and it acts like a phase-synchronized generator. —Quondum 02:11, 18 February 2014 (UTC)[reply]
The recent story about José Salvador Alvarenga made me wonder how long humans can survive at sea. Is it possible for a small group of men and women to survive for one year in a canoe from 12,000 years ago? (I think this is long enough to drift directly from Asia to the Americas, but I'm not sure.) Could these prehistoric boats survive the Pacific ocean, even if their occupants can't?
Of course we already know that most Native Americans came by land through Alaska. I'm just curious about whether involuntary trans-Oceanic trips were possible with Ice Age human technology. --140.180.246.0 (talk) 18:13, 10 February 2014 (UTC)[reply]
Pre-Columbian contact is not that unlikely, but it probably wouldn't have been very long before Columbus, for example 1200 AD. And there are many possible ways of verifying such contact, for example finding a native American skeleton from the pre-Columbian era that has a Polynesian component in its genome. Looie496 (talk) 18:32, 10 February 2014 (UTC)[reply]
Yes, I was conflating time periods in my answer. Also, I didn't mean that such evidence was conceptually difficult to verify, just that it would be a proverbial needle in a haystack. Though such a skeleton would be good, hard, evidence, AFAIK none have been found. SemanticMantis (talk) 20:27, 10 February 2014 (UTC)[reply]
(ec) It would depend on the kind of boat they were able to build, something we don't know much about. But I think it is unlikely, given that even the Polynesians, with their sophisticated sailing technology, didn't make it to Hawaii and Easter Island until sometime in the first millennium AD. Looie496 (talk) 18:28, 10 February 2014 (UTC)[reply]
Hawaiʻiloa, a double-hull sailing canoe built as a replica of Polynesian voyaging canoes.
As far as how long somebody could survive in an open canoe, I would think there is no set limit. Certainly the odds of them surviving would go down with time, but, provided they didn't hit a big storm, were able to get food from fishing and water from rain, and they stay in the tropics so they don't freeze to death, I don't see any critical resource which they will run out of. Some type of cover for the canoe so it could shed rain instead of flood would be important, and it would also prevent severe sunburns. StuRat (talk) 01:24, 11 February 2014 (UTC)[reply]
There is a lot of difference between people deliberately setting out to live on the ocean for long periods and the Alvarenga case. He was washed away from shore during a one day fishing trip - he was completely unprepared, yet managed to survive for over a year at sea. But if he'd been able to prepare, he'd have been much better off - and presumably would have been able to survive in much better condition. But if he could survive for a year - he could probably have survived almost indefinitely. If you imagine a fleet of small canoes - prepared for the trip - with small lateen sails to help them get where they needed to go - with people who spent most of their lives in canoes and tons of experience navigating by the stars...it seem very possible for them to cross the pacific. The Kon-Tiki expedition (and the MANY efforts to repeat it) showed that it's at least plausible for primitive peoples to survive very long distance voyages. SteveBaker (talk) 16:08, 11 February 2014 (UTC)[reply]
But again we should emphasize that while it is possible to survive indefinitely, it's not likely over such long periods, due to storms, holes or cracks in the canoe hulls, etc. Luck would need to be on their side. StuRat (talk) 16:34, 11 February 2014 (UTC)[reply]
Polynesians didn't settle the pacific in little drifting canoes. They had very large double-hulled or outrigger canoes with sails on which they carried pigs, chicken and breadfruit with them. μηδείς (talk) 21:35, 11 February 2014 (UTC)[reply]
See also our article on Kon-Tiki, a 1947 experimental raft which aimed to prove that it was possible "that people from South America could have settled Polynesia in pre-Columbian times. Although most anthropologists as of 2010 had come to the conclusion they did not..." However, six people "sailed the raft for 101 days over 6900 km (4,300 miles) across the Pacific Ocean before smashing into a reef at Raroia in the Tuamotu Islands". Alansplodge (talk) 19:08, 13 February 2014 (UTC)[reply]
^Cite error: The named reference lamptech was invoked but never defined (see the help page).
