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October 25

Daylight saving time

Thank you for your article on the above. Could you explain to me why there is such a great discrepancy in the dates of starting and ending DST?

In Ireland 2015, DST started on 29/3/2015 and ended 25/10/2015.

The start date in March is 119 days after the winter solstice (21/12/2014). The end date in October is 57 days before the winter solstice 21/12/2015.

I assume the Earth travels around the sun going to and away from the solstice at roughly the same speed and that daylight length is roughly equal for each corresponding day either side of the solstice. Why then the discrepancy?

Should not the two dates be an equal number of days away from the solstice?

Many thanksUnseenUni (talk) 10:08, 25 October 2015 (UTC)[reply]

The short answer is that in many cases, DST decisions are based on politics, not logic or science. —Steve Summit (talk) 11:31, 25 October 2015 (UTC)[reply]

There is significant logic to the concept of commencing daylight saving on or after the vernal equinox; and ending on or before the autumnal equinox. However, it is too simplistic to imagine that the weather around the vernal equinox is similar to that around the autumnal. At the time of the vernal equinox the whole hemisphere is still showing the last of the effects of winter so the weather is generally colder than around the time of the autumnal equinox when the hemisphere is still showing the last effects of summer. This consideration is relevant when decisions are made about the start and finish dates of DST. Dolphin (t) 11:40, 25 October 2015 (UTC)[reply]

You're making it sound like there's science behind it. Try candy. 99.235.223.170 (talk) 13:18, 25 October 2015 (UTC)[reply]

Heck, even the time zones that serve as a starting point for the need for DST are politically motivated.[1] - so no, it's definitely not science. SteveBaker (talk) 14:30, 25 October 2015 (UTC)[reply]

It's certainly not a coincidence that the US'll be moving the clocks back in the wee hours of Nov. 1st, right after trick-or-treating ends. μηδείς (talk) 16:04, 25 October 2015 (UTC)[reply]
- Not too many years ago, it was the Sunday before Halloween. Presumed to be safer for the kids this way. ←Baseball Bugs ^{What's up, Doc?} carrots→ 16:24, 25 October 2015 (UTC)[reply]
  - Yes, we never started trick-or-treating until it got dark, but nowadays trick-or-treating is pretty much over by the time it gets dark. μηδείς (talk) 17:05, 25 October 2015 (UTC)[reply]
    - Maybe that was in the days before child-haters started putting razor blades in the apples, and such stuff as that. ←Baseball Bugs ^{What's up, Doc?} carrots→ 18:47, 25 October 2015 (UTC)[reply]
      - Does it harm a kid less if he bites a razor blade before it's dark? --3dcaddy (talk) 00:33, 26 October 2015 (UTC)[reply]

Baseball Bugs. I suggest you read Poisoned candy myths. And if you already knew that it was a myth then you should have said so. CambridgeBayWeather, Uqaqtuq (talk), Sunasuttuq 00:26, 26 October 2015 (UTC)[reply]

CambridgeBayWeather I do not believe it's a myth. Do not believe everything you read in wikipedia. Do believe everything you read in snopes, which claims that some tampering of the trick-or-treat loot with razor blades and the like is documented. --3dcaddy (talk) 00:34, 26 October 2015 (UTC)[reply]

And note, in the Snopes page, that there were news reports of this sort of tampering as early as the 1960s... decades before DST in the US and Canada was extended past Halloween. --70.49.170.168 (talk) 05:40, 26 October 2015 (UTC)[reply]

The DST time change is synchronised throughout the EU, and occurs at 01:00 UTC on the last Sunday in March, and and 01:00 UTC on the last Sunday in October. See Summer_Time_in_Europe for the details. I don't know why those dates were chosen, but Sunday morning is the least disruptive time to change the clocks. — Preceding unsigned comment added by 62.56.61.161 (talk) 19:35, 25 October 2015 (UTC)[reply]

To address the OP's –

"I assume the Earth travels around the sun going to and away from the solstice at roughly the same speed and that daylight length is roughly equal for each corresponding day either side of the solstice. Why then the discrepancy?"

– this is not entirely true, because Perihelion occurs in January, so the earth is travelling slightly faster in its orbit between the Autumn and Spring equinoxes than it is between the Spring and Winter Equinoxes. See Equation of time. {The poster formerly known as 87.81.230.195} 185.74.232.130 (talk) 14:45, 26 October 2015 (UTC)[reply]

Thanks for the replies. I'm not sure what to make of the razor-blades, trick or treat, avenue of discussion so I'll leave that alone. However in response to the weather and temperature argument (Dolphin 51) surely the only reason for the change is the length of useable light in each day NOT the temperature. If there is enough light 53 days before the solstice why do we have to wait 119 days after the solstice before we see the light again (so to speak)?UnseenUni (talk) 19:07, 27 October 2015 (UTC)[reply]

Daylight savings time in the United States starts earlier and ends later than any of the places mentioned yet (second Sunday in March till the first Sunday of November). It says the change to November was made so there's more time to trick or treat before dark. It doesn't say why it's March 8-14 but if they made it even there would be DST in February and maybe it's too weird to change the clocks for the summer in a blizzard so they compromised and made it March (usually the November clock change is warmer than the March one, except maybe in a very continental part of America). Sagittarian Milky Way (talk) 00:19, 28 October 2015 (UTC)[reply]

for faster cooling, leave in open air or put object in palm

OP curious, cooling things, only these optionsMahfuzur rahman shourov (talk) 14:33, 25 October 2015 (UTC)[reply]

If it's a pot of boiling water, definitely leave it on the stove rather than putting it in your hand. In general, unless the air temperature is 98.6 degrees, leaving it in open air with circulation on all sides of it (such as a cooling rack) should be faster than in your hand. ←Baseball Bugs ^{What's up, Doc?} carrots→ 14:40, 25 October 2015 (UTC)[reply]

@Baseball Bugs:air is in 20-35 celsius rangeMahfuzur rahman shourov (talk) 14:51, 25 October 2015 (UTC)[reply]

Sure. The point being that if the air is warmer than your hand, leaving it in open air might not be as fast, but it might be safer. If the object is very much warmer than your hand is, it could cause serious damage. ←Baseball Bugs ^{What's up, Doc?} carrots→ 15:00, 25 October 2015 (UTC)[reply]

Let's remove some variables. Given an object that is not too hot to hold and small enough to fit in your hand, and given your hands are some temperature around 36c, if the air is at the SAME temperature as your hands, your hands will cool the object down faster. Water is a much better conductor of heat than air. This is why you can stick your arm into a 100c oven, but you couldn't stick your arm into a 100c pot of water. Conversely, you can easily hold your arm in a fridge that is 4c but holding your arm in 4c water would become very uncomfortable very quickly, the water is "cooling down" your arm much faster than the air. The fleshy parts of your hand is 70-80% water and your circulatory system conducts at least part of the heat away from your hand, so hand wins. However if the air is colder than your hand, or if the air is moving quickly, like a fan over a heat sink, this will affect the results considerably. At some point when the air is cold enough or moving enough the air will win, there is no universal answer. Vespine (talk) 01:21, 26 October 2015 (UTC)[reply]

Note that there are other factors at work besides the temperature of the "sink" (air or hand):

1) The density of the material also matters. A more dense material is able to hold more heat, and thus draw away more heat. The density of the human hand is roughly that of water, which is far higher than air.

2) Circulation also matters. If there is no wind, circulation in air may still occur due to convection, but this is rather dependent on the shape of the container. As for circulation in the hand, this is driven by the heart and circulatory system. So, which would have faster circulation is hard to say. StuRat (talk) 01:17, 26 October 2015 (UTC)[reply]

I see your point. It would also be helpful if the OP would tell us just what kind of object he has in mind, and how hot it is. ←Baseball Bugs ^{What's up, Doc?} carrots→ 01:47, 26 October 2015 (UTC)[reply]

Calorie / Energy requirements men vs women

Why is it men always have a higher nutritional requirement than women. Surely, a man and woman of equal statute require the same level of food? — Preceding unsigned comment added by 80.195.27.47 (talk) 17:23, 25 October 2015 (UTC)[reply]

testosterone production by testicles, resulting anabolismMahfuzur rahman shourov (talk) 17:40, 25 October 2015 (UTC)[reply]

If by "nutritional requirement" you mean calorie intake, men tend to be taller and more massive, due to the anabolic effects of testosterone, so, on average, men will burn more calories. Of course, when comparing individuals, rather than looking at population statistics, there will be a lot of individual variation. Certainly, women require more calories than they otherwise would when pregnant or lactating. --71.119.131.184 (talk) 17:48, 25 October 2015 (UTC)[reply]

Re: "men always have a higher nutritional requirement than women", I believe this is an incorrect statement. For example, a male midget presumably needs fewer nutrients than a female giant. As for two of the same mass, I'm not sure this statement is true, then, either, although men tend to have more muscle and women have more fat, and muscle burns more calories than fat. StuRat (talk) 01:21, 26 October 2015 (UTC)[reply]

This is just a very rough guideline. It's quite easy to find some articles discussing this very subject. In short, the 1st IP's reply pretty much sums it up. Vespine (talk) 04:29, 26 October 2015 (UTC)[reply]

@Vespine:first ip reply same mine,i post beforeMahfuzur rahman shourov (talk) 15:08, 26 October 2015 (UTC)[reply]

Power → acceleration at low speeds

Hello all. Trying to calculate vehicle acceleration and top-speed, based on engine power output. Traditional formula: power = mass × acceleration × velocity. But that seems to imply that at v=0, any power gives infinite acceleration?

Turning to the formula for kinetic energy, the kinetic energy = 0.5 × mass × v². So if during a given period of power usage the kinetic energy has changed by δKE, then velocity after that timestep is v2 = sqrt(v1² + (dt × 2 × δKE / mass)). Does this work? What are the units? 146.198.171.0 (talk) 19:36, 25 October 2015 (UTC)[reply]

You may find Zeno's paradoxes to be somewhat relevant. ←Baseball Bugs ^{What's up, Doc?} carrots→ 20:11, 25 October 2015 (UTC)[reply]

Indeed, or just a formula that allows calculating the "velocity after x seconds from a standing start" without having to use tiny timesteps to get sufficient accuracy... 146.198.171.0 (talk) 20:20, 25 October 2015 (UTC)[reply]

You can just use:

v={\sqrt {{2Pt} \over {m}}}

where

v

is the velocity,

P

is the power,

t

is the time, and

m

is the mass. The energy supplied to the body is just Power x Time, and the velocity can be calculated using the formula for kinetic energy. Tevildo (talk) 21:46, 25 October 2015 (UTC)[reply]

v={\sqrt[{2}]{2Pt \over m}}

and

a={\sqrt[{2}]{P \over 2mt}}

units are m/s, W, s, kg, m/s²

But to know the acceleration at low speeds, you have to know the power at low speeds, and that will be a function of speed...

Usually the torque, not the power is used: an engine can produce torque at v=0, but not power, because as you say, that would result in an infinite acceleration. The torque an engine can produce is limited, and the power equals torque*angular_speed, so at small speeds and with torque limited, you won't be able to use max power. A typical curve is found here With only power given, I guess you can use the formula above, but realize that the resulting acceleration at low speeds is not realistic. Ssscienccce (talk) 21:44, 25 October 2015 (UTC)[reply]

I see Tevildo was faster... weird, my time stamps is 2 minutes before his, but when I saved his post appeared... Ssscienccce (talk) 21:50, 25 October 2015 (UTC)[reply]

Same here. ;) We must have _just_ missed an edit conflict. Tevildo (talk) 22:03, 25 October 2015 (UTC)[reply]

And even power & torque curves for an electric motor, such as this one (for various versions of the Tesla Model S), show zero power at zero speed, though they do have their best torque at low (including zero) speeds. -- ToE 03:19, 26 October 2015 (UTC)[reply]

The original question essentially stems from the commission of a division by zero error. The formula,

P=a\times v\times t

, contains values (acceleration, velocity, and time) that are related to each other by differentiation. They are not ordinary numbers. When you attempt to solve for power at a velocity of zero, you must account for the fact that instantaneous application of a differential element of acceleration will instantaneously and continuously affect that velocity. Essentially, you are seeing a solution where an infinite amount of power is applied for an infinitely short amount of time, and your intuition is that this is somehow meaningful ... but it's not, because not all infinities are equivalent when you compare their sizes. This is a technique of calculus; you cannot simply treat the values in this equation as regular constant numbers and solve algebraically. The simple formula only works as a steady-state simplification of a more complete treatment of the kinematics. The power applied when velocity is exactly zero is not a meaningful quantity: power is a derivative of total energy.

Using the same division-by-zero error, we can manipulate the equation

P=a\times v\times t

to incorrectly demonstrate that at zero velocity, power is zero, power is "one," (in arbitrary units), power is infinite... the magic of wrong math is that it can rapidly become very wrong!

When we try to teach physics without calculus, some books simplify this by considering average power, which is rate of change of energy per unit of time, averaged over a non-zero amount of time. This lets students solve simplified problems using only the techniques of algebraic equation manipulation. But, if you use this simplified variation, you must not try to take the equation to the limit as the time interval goes to zero: we have an entire branch of math that explains how to construct that problem more effectively. Nimur (talk) 22:19, 25 October 2015 (UTC)[reply]

Nimur, are you saying that there exists a differentiable function v(t) with v(t₀)=0 for which p(t₀)≠0 when we define p=d/dt(0.5·m·v²) for a fixed value m? -- ToE 02:59, 26 October 2015 (UTC)[reply]

... the point being that there is no such function, and the questioner was right in seeing infinite acceleration for non-zero power at zero velocity as a red flag indicating that something was wrong with their approach. The problem was not a division by zero error as much as a misunderstanding of the physics and engineering. As Tardis points out below, while the engine might be generating instantaneous power while the vehicle is not moving, that power is fully lost (via transmission or clutch) during the instant of zero velocity. I mentioned the power & torque curve for an electric vehicle above. For an internal combustion engine, the shaft horsepower curve will not extend below the idle RPM of the engine, but the wheel horsepower curve of the vehicle will be zero at zero velocity. Having 300 ponies under the hood does not mean that the engine will be able to do 300 hp of work on the vehicle off the line. -- ToE 19:27, 29 October 2015 (UTC)[reply]

The power you're using there is the change of KE of the vehicle over time, not the engine's power. What you've discovered is that the car's efficiency is instantaneously 0 when it is stopped. This should come as no surprise: it's certainly 0 while sitting still, and it's not unreasonable that it should increase continuously therefrom as you accelerate. --Tardis (talk) 16:02, 26 October 2015 (UTC)[reply]

Condoms

Who do condoms only cover the shaft of the penis and not the scrotum or areas around it? 2A02:C7D:B8FF:7E00:C1D3:14DB:4194:8D14 (talk) 19:56, 25 October 2015 (UTC)[reply]

Misinformation/Joke

That's not always true.[2] — Preceding unsigned comment added by Baseball Bugs (talk • contribs) 20:34, 25 October 2015 (UTC)[reply]

WP:NOTAFORUM Try to bear in mind that if someone does not (or cannot) follow up your link, they may very well go away with misinformation about an important topic, taking your comment at face value. Regardless, there are plenty of places on the internet where you can reference your favourite comedies to your heart's content; WP really should not be one of them. Snow ^{let's rap} 21:55, 25 October 2015 (UTC)[reply]

WP:NOTGETTINGIT Bestfaith (talk) 19:19, 28 October 2015 (UTC)[reply]

On the contrary, wearing a rubber suit should afford great protection. ←Baseball Bugs ^{What's up, Doc?} carrots→ 01:38, 26 October 2015 (UTC)[reply]

I don't think you read what I wrote. Snow ^{let's rap} 02:10, 26 October 2015 (UTC)[reply]

I don't think you watched the video. And calling it "misinformation" is silly - as well as being untrue. ←Baseball Bugs ^{What's up, Doc?} carrots→ 15:48, 26 October 2015 (UTC)[reply]

The reason that condoms can help prevent infection for many (though by no means all!) STD's without covering the entire surface of the male genitalia is two-fold. First, this typically suffices to capture the fluids produced by the penis during copulation which are often the medium through which infection can be spread; this obviously protects the man's partner. But the condom also protects the person wearing it by covering the glans, meatus, and (in the uncircumcized), the foreskin; the surface tissues of these structures contain specialized cells which are particularly vulnerable to the pathogens responsible for some STDs. It also helps prevent fluids from entering back into the urethra, to similar effect. These protective measures are not, of course, 100% effective, nor do they prevent against all forms of STD, but they do help markedly with regard to some of the most vulnerable tissues and some of the worst diseases. Snow ^{let's rap} 21:55, 25 October 2015 (UTC)[reply]

Note that the scrotum is important for thermal control of the temperature of the testicles, which in turn is important for sperm production. Of course, if he is using a condom you might assume he doesn't care if the current batch of sperm live or die, but nature has also provided men with an aversion to any temperature which causes sperm to die, so having the testicles overheat would completely ruin the experience for most men. StuRat (talk) 01:26, 26 October 2015 (UTC)[reply]

I see. Then why do many people not use condoms for oral sex whether cunnilingus or fellatio? Surely that's just as risky as for example saliva or other fluids could come into contact with those vulnerable surface tissues you mentioned. — Preceding unsigned comment added by 176.24.24.18 (talk) 07:36, 26 October 2015 (UTC)[reply]

Using a condom during fellatio is definitely recommended, particularly during high risk encounters (with prostitutes for example). Using a cut out condom is one option, but a dental dam is much preferred during cunnilingus. See oral sex and safe sex for more information. As to why some people don't use such recommend protections, people make their own decisions based on a whole host of facts and STI risk is only one thing they may consider, and many people may not even properly understand the risks anyway. After all, some men receive bareback anal sex from a male prostitute, something which would generally be considered a really high risk activity. And the problems married couples, particularly a wife faces trying to negotiate condom usage with her husband, both as a contraceptive and for protection against STIs in some regions with high rates of HIV are commonly discussed [3]. Nil Einne (talk) 12:02, 26 October 2015 (UTC)[reply]