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April 10

Special relativity. Simultaneity 2.

https://en.wikipedia.org/w/index.php?title=Wikipedia:Reference_desk/Science&action=edit&oldid=714473425

Question

Remark

You said the hind clock, when it shows the numerical value $\theta _{0}^{[}$ on its face, is at $t'=\theta _{0}^{[}$. It was not statement , but assumption as I'm not sure it's true (even for previous thread). I've taken it from previous thread (see Remark). So, we shall heve next situation: ε' frame ε frame time coordinate clock reading time coordinate clock reading $t'=0$ $\theta _{0,self}^{[}=-{\tfrac {L'}{2c}}$ $\theta _{0,self}^{]}=-{\tfrac {L'}{2c}}$ $t=0$ $\theta _{0}^{[}=\theta _{1}^{[}-\tau _{1}{\sqrt {...}}$ $=-\tau _{1}{\sqrt {1-u^{2}/c^{2}}}$ $\theta _{0}^{]}=\theta _{2}^{]}-\tau _{2}{\sqrt {...}}$ $=-\tau _{2}{\sqrt {1-u^{2}/c^{2}}}$ $t'=+{\tfrac {L'}{2c}}$ $\theta _{1,self}^{[}=\theta _{2,self}^{[}=0$ $\theta _{2,self}^{]}=\theta _{1,self}^{]}=0$ $t=\tau _{1}={\tfrac {L/2}{c+u}}$ $\theta _{1}^{[}=0$ $t=\tau _{2}={\tfrac {L/2}{c-u}}$ $\theta _{2}^{]}=0$ $t'=\theta _{0}^{[}+{\tfrac {L'}{2c}}$ $\theta _{3,self}^{[}=\theta _{0}^{[}$ $\theta _{3,self}^{]}=\theta _{0}^{[}$ ... $t'=A$ $\theta _{(t'=A),self}^{[}=A-{\tfrac {L'}{2c}}$ $\theta _{(t'=A),self}^{]}=A-{\tfrac {L'}{2c}}$ $t=A$ $\theta _{(t=A)}^{[}=(A-\tau _{1}){\sqrt {..}}$ $\theta _{(t=A)}^{]}=(A-\tau _{2}){\sqrt {..}}$ Correct? Let's check. Hind clock at start is situated in ε in point $(x,t)=(-L/2,0)$ and in ε' in point $(x',t')=(-L'/2,0)$. $(x',t')=({\tfrac {x-ut}{\sqrt {1-u^{2}/c^{2}}}},{\tfrac {t-xu/c^{2}}{\sqrt {1-u^{2}/c^{2}}}})$; $(x,t)=({\tfrac {x'+ut'}{\sqrt {1-u^{2}/c^{2}}}},{\tfrac {t'+x'u/c^{2}}{\sqrt {1-u^{2}/c^{2}}}})$. What's next? What to substitute in and what should be compared with result?

https://en.wikipedia.org/w/index.php?title=Wikipedia:Reference_desk/Science&action=edit&oldid=713271421 Special relativity. Simultaneity. In your problem, say the light is emitted at (0,0) (in either coordinate system). It reaches the walls at $(x',t')=(\pm L/2,L/2c)$, or equivalently $(x,t)=({\tfrac {\pm Lc}{2(c\pm u)}},{\tfrac {L}{2(c\pm u)}})$. Your first clock has the parametric equation (x,t)(τ) = (0,τ), which you can plug into the Lorentz transformation to find (x',t')(τ). To find the digital reading of that clock at a particular (x,t) or (x',t'), you just solve for τ. Your other three clocks have the parametric equations (x',t')(θ) = (0, θ) and (x',t')(θ[]) = (±L/2, θ[]), and you can plug those into the Lorentz transformation to find (x,t)(θwhatever). Any question you have about what someone sees at a given (place,time) can be answered in this way. They are simple questions about coordinate geometry. -- BenRG (talk) 19:58, 29 March 2016 (UTC)

37.53.235.112 (talk) 12:02, 8 April 2016 (UTC)

OK. According BenRG's suggestions we have:

$(x';t')(A)=(-{\tfrac {L'}{2}};A-{\tfrac {L'}{2c}})$;

$(x,t)=({\tfrac {x'+ut'}{\sqrt {1-u^{2}/c^{2}}}},{\tfrac {t'+x'u/c^{2}}{\sqrt {1-u^{2}/c^{2}}}})$;

we will substitute x' and t' and should get next

$\color {red}{(x;t)(A)=(-{\tfrac {L}{2}}+uA;(A-\tau _{1}){\sqrt {1-u^{2}/c^{2}}})}$.

After substitution:

$(x,t)(A)=({\tfrac {(-{\tfrac {L'}{2}})+u(A-{\tfrac {L'}{2c}})}{\sqrt {1-u^{2}/c^{2}}}},{\tfrac {(A-{\tfrac {L'}{2c}})+(-{\tfrac {L'}{2}})u/c^{2}}{\sqrt {1-u^{2}/c^{2}}}})$;

$(x,t)(A)=({\tfrac {uA}{\sqrt {1-u^{2}/c^{2}}}}-{\tfrac {L/2}{1-u/c}},{\tfrac {A}{\sqrt {1-u^{2}/c^{2}}}}-{\tfrac {L/2}{c+u}})$.

Why result doesn't match with highlighted equation? 37.53.235.112 (talk) 18:02, 9 April 2016 (UTC)

There's a sign error in the first equation (below the box); it should be $(x';t')(A)=(-{\tfrac {L'}{2}};A+{\tfrac {L'}{2c}})$. With that change, the Lorentz transformation should get you $(x,t)(A)=(u\gamma A-c\tau _{1},\;\gamma A+\tau _{1})$, where I've introduced $\gamma =1/{\sqrt {1-u^{2}/c^{2}}}$ and used your definition of $\tau _{1}$. This is close to the answer you got, and I think the difference is because of the sign error. You should be able to look at this answer and see that it makes sense. For example, if you plug in $A=0$, you get $(-c\tau _{1},\tau _{1})$, which is where the light reaches the clock. Also, the factor of $u$ gives the clock the correct speed, and the factor of $\gamma$ gives it the correct time dilation factor.

Your expected answer (in red) is wrong in several ways. Again, try plugging in $A=0$, think about whether the velocity of the clock (the slope of the line in spacetime) is $u$, and think about whether you should multiply or divide by the time dilation factor. -- BenRG (talk) 05:33, 10 April 2016 (UTC)

Is table ε' frame — ε frame correct? 37.53.235.112 (talk) 07:22, 10 April 2016 (UTC)

Yes, I think so. (Note that where you have $(A-\tau _{1}){\sqrt {..}}$ in that table, it's a clock reading (correct), whereas in your prediction in red, it's a $t$ coordinate (incorrect).) -- BenRG (talk) 07:30, 10 April 2016 (UTC)

How to check table ε' frame — ε frame through Lorentz transformations? Namely, is there correlation between left (ε') and rigth part (ε)? 37.53.235.112 (talk) 09:06, 10 April 2016 (UTC)

The Lorentz transformation will turn the $(x,t)$ of a clock reading into the $(x',t')$ of the same clock reading, or vice versa. In your table you don't have any $x$ or $x'$, and most of the clock readings on the left aren't on the right and vice versa. You could add $x$ and $x'$ coordinates for the points where the clocks read $0$, then check that they match using the Lorentz transformation. But it makes more sense to derive a general equation for the $(x,t)$ and $(x',t')$ coordinates of the clocks as a function of their reading. You just did that for the hind clock (albeit with some errors). Then you can forget about the table, which just gives the values of those functions at certain points. -- BenRG (talk) 17:32, 10 April 2016 (UTC)

Why are images deleted? E.g.

https://en.wikipedia.org/w/index.php?title=Wikipedia:Reference_desk/Science&action=edit&oldid=713271421 Special relativity. Simultaneity

https://upload.wikimedia.org/wikipedia/commons/1/17/160329155625UTC.png

https://upload.wikimedia.org/wikipedia/commons/4/49/160329181132UTC.PNG


https://en.wikipedia.org/w/index.php?title=Wikipedia:Reference_desk/Science&action=edit&oldid=714473425 Special relativity. Simultaneity 2

https://upload.wikimedia.org/wikipedia/commons/8/8f/Reference_desk_Science160404140000.PNG

Whati is expiration time of Wikipedia Upload Wizard? Where is this info in rules?..

OP see ]https://commons.wikimedia.org/wiki/Commons:Deletion_requests/Files_uploaded_by_Image_upload_160329142034UTC]. I don't know the details of the image deletions nor Commons policies. --Modocc (talk) 14:36, 14 April 2016 (UTC)

Help identify this plant, please

{ Moved from Miscellaneous Ref Desk }

A plant similar to Calycopteris floribunda but with different kind of flowers

A plant similar to Calycopteris floribunda but with different kind of flowers

A climber closely similar to Calycopteris floribunda but never seen to grow as big as that. The plant grows in Kerala, India and flowers during the heights of hard summer. Thrives in wild and don't seem to require watering at all for the flowering. Flowers are fragrant, though not intensely. What is it? — Preceding unsigned comment added by 117.253.195.109 (talk) 02:18, 10 April 2016 (UTC)

The hanging clusters of blossoms opening downward and the leaves resemble Hoya carnosa (familiar name: wax plant, porcelain flower). It's familiar to me as a shrub, possibly cultivated (i.e. not seen in the wild), in the arid subtropic climate zone. -- Deborahjay (talk) 07:34, 10 April 2016 (UTC)

Hoya carnosa is certainly is not by the photos I see on its page. The woody vine and leaves are very similar to Calycopteris floribunda but the flowers are not. --117.253.195.109 (talk) 08:20, 10 April 2016 (UTC)

To me this looks like Combretum indicum, or a closely related species of Quisqualis or Combretum. --Dr Dima (talk) 20:47, 11 April 2016 (UTC)

2016 Mercedes GLE 350d urea injection

My friend says that the 2016 Mercedes GLE 350d doesn't require urea fill-ups anymore unlike the previous version and most diesel cars on the market. Is this actual true?

Our article Mercedes-Benz_M-Class#Engines says "ML 350 BlueTEC 4MATIC", and BlueTEC is Mercedes' branding of urea injection, so just based on the model number I'm inclined to say he's wrong, but I can't find any sources to back me up. Johnson&Johnson&Son (talk) 10:42, 10 April 2016 (UTC)

Ok， so I found three sources[1][2][3] saying that there was a renaming of the model from "ML 350 BlueTEC 4MATIC" to "GLE 350d 4MATIC" back in 2015. Removing "BlueTEC" from the model name could count as circumstantial evidence.

I also found this Mercedes Australia page[4] which names 4 models that require urea injection and the GLE 350d 4MATIC isn't among the four. But this is circumstantial evidence again since it could simply be that the page is outdated or that they don't sell that model in Australia. Johnson&Johnson&Son (talk) 11:12, 10 April 2016 (UTC)

I added a redirect from urea injection to diesel exhaust fluid. StuRat (talk) 14:26, 10 April 2016 (UTC)

This which is about the 2016 Mercedes-Benz GLE 350d 4Matic Coupe [5] says "It involves automatically injecting a liquid called AdBlue into the exhaust stream at pre-determined intervals, breaking the harmful nitrogen oxides into harmless water and nitrogen." Nil Einne (talk) 16:04, 10 April 2016 (UTC)

This description of a very lightly used 2016 Diesel GLE 350d 4Matic AMG Line 5dr 9G-Tronic Auto [6] mentions "BlueTEC diesel emission control system including AdBlue reservoir". The current description doesn't mention that but has also been changed in other ways. This [7] says "The diesel GLE uses injections of AdBlue to reduce harmful emissions" although weirdly it doesn't talk about 350d despite being about the 2016 models. It does mention the GLE300d 4MATIC. This [8] isn't in English but despite not mentioning 2016, from the date and names appears to be about the 2016 models and you can see AdBlue is indicated even without machine translation. This [9] also isn't English but if you scroll right to the 4th of 6th photos at the top it shows an AdBlue tank and the headline machine translates to something like "AdBlue, the only solution for a less harmful diesel". This [10] also isn't in English but a machine translation says something like "Although already new signage does not know about it (déčko is replaced by the CDI, or with a BlueTEC) next to the conventional 93-litre fuel tank is also traditionally the urea tank (AdBlue liquid) that is used for injection into the exhaust system-most of the NOx into nitrogen and water changes". Nil Einne (talk) 16:47, 10 April 2016 (UTC)

The above used car which mentions the manuala got me thinking so I searched for Mercedes manuals and found [11] which is supposed to have 2016 manuals. Looking in the GLE section, there's a a link to [12] which has a filling capacity for the GLE 350 d 4MATIC Coupe for DEF (28.4L). It also links to [13] which has the same filling capacity for the GLE 250 d 4MATIC Sport Utility Vehicle and GLE 350 d 4MATICSport Utility Vehicle. Nil Einne (talk) 17:05, 10 April 2016 (UTC)

Many thanks!!!Johnson&Johnson&Son (talk) 02:41, 11 April 2016 (UTC)

Glad to help. Nil Einne (talk) 14:51, 11 April 2016 (UTC)

Resolved

Sunlight in June

The introduction to June claims that it is the month with greatest amount of daylight, in the Northern Hemisphere. Given the slow per-day changes in amount of daylight around the times of the solstices, how is June the correct answer? May and July both have an additional day, and July in particular is closer to the solstice than May, so how is it that July doesn't have the greatest amount of daylight? Of course, it makes sense for June to have the shortest amount of daylight in the Southern Hemisphere, since it has the winter solstice and one day fewer than either adjoining month. Nyttend (talk) 15:40, 10 April 2016 (UTC)

The quote from our article is "June is the month with the longest daylight hours of the year in the Northern Hemisphere and the shortest daylight hours of the year in the Southern Hemisphere". It's ambiguous if that means per day or the total for the entire month. And if if it does mean per month, there's only 3% more days in July, so even a minor difference per day could be more than that. Also note that how much the length of the day changes from June to July varies with the distance from the equator, so the answer might be different in different locations. StuRat (talk) 16:00, 10 April 2016 (UTC)

(ec)Looks to me like June would be correct. About three-fourths of the longest days of the season occur in June, and the fact of May and June having extra days does not make up for the difference...

The summer solstice typically occurs on June 21, but this year it's the 20th, this being a leap year. Checking my Old Farmer's Almanac for 2016 (which uses Boston, MA as its reference point, about 42 degrees north latitude):

May 1 has only 14 hours and 5 minutes of sunlight. Its first day with 15:00 or more sunlight is the 28th.

June 1 has 15:05 sunlight, peaking at 15:18 on the 20th and 21st, and is 15:14 on the 30th.

July 1 has 15:14, but its last day with 15:00 or more of sunlight is the 14th. The 31st is 14:28.

If the solstice occurred at midnight between June 30 and July 1, then July would have the most daylight, by 1 day. But because it's June 20 or 21, June wins. ←Baseball Bugs ^{What's up, Doc?} carrots→ 16:21, 10 April 2016 (UTC)

No, I've got it wrong. See calc's below. ←Baseball Bugs ^{What's up, Doc?} carrots→ 17:30, 10 April 2016 (UTC)

Let's do some math with those numbers, just using linear interpolation (hopefully close enough):

May average = {14 hrs 5 mins (14.083) + 15 hrs 3 mins (15.050)} / 2 = 14.567
May daylight hours = 31 × 14.567 = 451.577 hours 

July average = {15 hrs 14 mins (15.233) + 14 hrs 28 mins (14.467)} / 2 = 14.850
July daylight hours = 31 × 14.85 = 460.35 hours

June average, up to solstice =  {15 hrs 5 mins (15.083) + 15 hrs 18 mins (15.300)} / 2 = 15.192
June daylight hours, up to solstice = 21 × 15.192 = 319.025 hours
June average, after to solstice =  {15 hrs 18 mins (15.300) + 15 hrs 14 mins (15.233)} / 2 = 15.267
June daylight hours, after to solstice = 9 × 15.267 = 137.4 hours
June daylight hours = 319.025 + 137.4 hours = 456.425 hours

So, if my math is correct, it does look like July has more total daylight hours in Boston, in July, than in June. StuRat (talk) 16:46, 10 April 2016 (UTC)

[ec] On the other hand, this page (which defaults to the latitude of Ghent) gives a comprehensive list of times. After some work with Excel, it appears that Nyttend's suspicions are correct - the exact figures are: May, 483 hours 37 mins; June, 494 hours 14 mins; July, 495 hours 31 mins. Presumably there's a critical latitude (which varies from year to year) at which June and July are equal, but the unqualified assertion that June has the greatest amount of daylight is apparently not true. Tevildo (talk) 16:40, 10 April 2016 (UTC)

Technically two critical latitudes, as there is also a line near the north pole, above which both June and July have 24 hours of sunlight every day. StuRat (talk) 16:55, 10 April 2016 (UTC)

Using this site,[14] note that at Barrow, Alaska, which is above the Arctic Circle, every day in both June and July is listed as 24 hours of sunlight. So at the North Pole, July beats June by a full 24 hours.[15] ←Baseball Bugs ^{What's up, Doc?} carrots→ 17:44, 10 April 2016 (UTC)

Note further that in Quito, which is very close to the equator,[16] daylight is 12:07 all year, except for two weeks before and after the December solstice, when it's 12:08. So at the equator, July beats June, by 12 hours. ←Baseball Bugs ^{What's up, Doc?} carrots→ 18:13, 10 April 2016 (UTC)

But the article actually says "longest daylight hours", not "most hours of daylight". Stu says it's ambiguous; I say the use of "longest" obviously implies that it refers to the average daylight per day. It's still badly phrased, though. As noted, there are latitudes where June and July have continuous daylight. --69.159.61.172 (talk) 19:46, 10 April 2016 (UTC)

I reworded the article to be unambiguous and correct: "June contains the summer solstice, the day with the most daylight hours in the Northern Hemisphere and the fewest daylight hours in the Southern Hemisphere (excluding polar regions in both cases)." StuRat (talk) 20:56, 10 April 2016 (UTC)

You created a further ambiguity by mentioning only the summer solstice, so I have reworded as follows: "June contains the summer solstice in the Northern Hemisphere, the day with the most daylight hours, and the winter solstice in the Southern Hemisphere, the day with the fewest daylight hours (excluding polar regions in both cases)." Akld guy (talk) 22:08, 10 April 2016 (UTC)

Thanks. Even the Jarada would be proud of that. :-) StuRat (talk) 22:16, 10 April 2016 (UTC)

And I have just changed December along the same lines with the same wording. Akld guy (talk) 22:22, 10 April 2016 (UTC)

Unfortunately, "the day with the most daylight hours" implies that it is the only such day. As soon as you get a little way beyond the Arctic Circle, there is more than one day each year with 24-hour daylight. --69.159.61.172 (talk) 23:11, 10 April 2016 (UTC)

But the wording specifically excludes polar regions. Nyttend (talk) 00:04, 11 April 2016 (UTC)

Yes, I thought it was necessary to exclude polar regions, since they would require more explanation than readers would want to read. StuRat (talk) 02:45, 11 April 2016 (UTC)

User:Nyttend, I'm late to this party, but do check out Insolation and refs therein. No need to resort to OR and back of the envelope, this stuff is seriously well-studied. Using that keyword, lat/long, etc can get you very good data, even stuff that includes cloud models and other advanced features if you like. Here's a slightly older but fairly comprehensive overview [17]. SemanticMantis (talk) 14:17, 11 April 2016 (UTC)

It's pretty obvious that the statistic refers to the average amount of daylight per day - and not the total for the month...the latter would be a really useless statistic. SteveBaker (talk) 17:36, 11 April 2016 (UTC)

If that's the approach, June wins: 456.425 / 30 = 15.2 while 460.35 / 31 = 14.8. ←Baseball Bugs ^{What's up, Doc?} carrots→ 21:39, 11 April 2016 (UTC)

Criticality and Neutron flux

If I had two pieces of Pu(238) (each weighing 5Kg), one in each hand and brought them togther quickly, how much netron flux would I get and how far away would someone have to be to survive such a criticality incident?--178.101.224.162 (talk) 21:02, 10 April 2016 (UTC)

I think your critical mass for plutonium is off, but in any case see Demon core (which is too mild, come to think of it, since it didn't blow itself apart), and fizzle yield (which is too harsh, since people are trying to jam that together with more effort). Wnt (talk) 21:34, 10 April 2016 (UTC)

Critical mass is between 9 and 10 kg according to our article. So whats wrong with my figures?--178.101.224.162 (talk) 21:39, 10 April 2016 (UTC)

I know you aren't talking about a "weapon" per se, however I think most of the same "issues" would apply as described in this article: Nuclear_weapon_yield#Calculating_yields_and_controversy . Vespine (talk) 23:52, 10 April 2016 (UTC)

No. Not a weapon. Im talking about a criticality 'accident' where the main effect is a large neutron flux and maybe some gamma rays. Definitely not a nuclear explosion as we know it (Jim).--178.101.224.162 (talk) 00:23, 11 April 2016 (UTC)

yes I know, I said I know you aren't talking about a weapon, but my point is that it's very hard to predict, even with accurate models, precisely what will "happen", whether something will reach criticality or not or how much reaction will take place. I recall seeing an article about software that was used for such calculations and a lot of details about it were confidential, except that it was extremely complex, extremely expensive and extremely classified. Vespine (talk) 02:23, 11 April 2016 (UTC)

I stand corrected! I had read some much smaller figures for bombs .... never realized that the critical mass could be changed that much by design considerations. I shoulda RTFA! Wnt (talk) 02:08, 11 April 2016 (UTC)

Yeah I'm not considering use of Neutron reflection, although the water content of my body may have a small effect.--178.101.224.162 (talk) 13:30, 11 April 2016 (UTC)

Why do Electric aircraft have poor thrust to weight ratio?

Is it because of battery limitations or electric motor limitations? Also if electric vehicles have such poor thrust to weight ratios, why is it that the Tesla Model S has such a good 0-60 acceleration time? ScienceApe (talk) 23:30, 10 April 2016 (UTC)

Electric motors are lovely. Small, powerful and efficient. Batteries carry both the fuel and the oxidizer, as rockets do, and don't work on chemicals with as much energy per weight. Power is splendid; thus you can accelerate quicly but batteries are heavy and poor in energy, hence can't go far. Jim.henderson (talk) 23:36, 10 April 2016 (UTC)

Plus, the motors they use have fairly stable "speed-torque" curves. In other words, they give you pretty much the same force whether they're turning slowly or quickly, and they don't generally need a complicated gear/transmission system (a few older electric trains had them). Fuelled vehicles by contrast get the best torque within a relatively narrow range of speeds, and you need a complicated gearbox to get the most out of them. Where acceleration is your biggest concern, electric vehicles are great (look at List of trolleybus systems in the United Kingdom - most electric bus systems were either in very big cities where you have lots of stopping and starting, or in mountainous areas like Yorkshire and the Welsh Valleys. Because it doesn't have the extra weight of batteries, a trolleybus will just fly up a hill almost as if it was flat road). Smurrayinchester 08:06, 11 April 2016 (UTC)

(edit conflict) What makes you think that electric aircraft have poor thrust to weight ratio? I could build you an electric aircraft with a fansastic thrust to weight ratio. It might run the capacitors dead in the first 30 seconds, but until then it would take off like a bat out of hell. Now if you want a good thrust to weight ratio and a decent range, then our article on Energy density has some interesting news about gasoline vs. batteries vs,. capacitors... --Guy Macon (talk) 23:45, 10 April 2016 (UTC)

That's roughly what small electric quadcopters do. Nobody minds the short flights caused by their low energy to weight ratio. Big airplanes, people like to fly far and long. So, they load up with energy storing or making things. This converts the problem to one of power to weight. 108.14.112.81 (talk) 02:24, 11 April 2016 (UTC)

As a bit of interesting trivia, the aircraft with the largest thrust to weight ratios are all blimps or dirigibles. :) --Guy Macon (talk) 04:19, 11 April 2016 (UTC)

Use mass instead. 1989 Goodyear blimp plus pilot, little fuel: 14.4 lb/hp. 1998 world's fastest accelerating production sedan, full tank, empty trunk, four 157 lb men, 4 ¼ lbs clothes (x4): 14.5 lb/hp. City block-sized balloons with 840 horsepower is where it's at. Sagittarian Milky Way (talk) 07:23, 11 April 2016 (UTC)

As for the Tesla, weight isn't as much of a problem in a land vehicle, as you don't need to provide enough lift to offset it. StuRat (talk) 04:28, 11 April 2016 (UTC)

I think to directly answer the question then: battery weight for an airplane is a much more significant issue than it is for a car...but even electric car batteries don't last that long and are very expensive...the limitation is not electric motors themselves (they can be plenty powerful to create great thrust) but powering a powerful electric motor...your flight time could be a matter of minutes instead of hours...68.48.241.158 (talk) 16:48, 11 April 2016 (UTC)

note, too, your tesla car can do what you describe...but will run down the battery in no time...68.48.241.158 (talk) 16:58, 11 April 2016 (UTC)

Besides the huge advantage in fuelefficiency, tru there is way more power stored in 1 kg kerosene then can be loaded into 1 kg of battery, electricity can only use a Propeller (aeronautics), which is limited to many conditions and not very effective in general, to generate reasonable thrust while you have multiple options with fuel, including very powerfull such as turbines (with afterburners), which have the highest thrust to weight ratio of all powermachines (some even excel rockets (which are not machines:)). --Kharon (talk) 17:45, 11 April 2016 (UTC)

I presume we're talking about propeller driven planes here. Jets are different for lots of reasons.

One of the reasons why electric cars have such good acceleration is that electric motors have very flat torque curves - they provide more or less the same torque for any given motor speed below the maximum. Since torque and acceleration are closely related, they have very uniform acceleration at all speeds. So the 0 to 60mph time for a Tesla is almost identical to it's 60 to 120mph time. With fast gasoline powered cars, the torque curve is anything but flat - there is a relatively narrow band of RPM's at which they produce peak torque - and if they have turbochargers, getting uniform torque is even harder.

The (partial) fix for that in a car is to use lots of gears - but there are losses involved in a complex gearbox - and the act of shifting gears (with the requirement to match the engine RPM while shifting) takes time and results in the engine RPM going in and out of the "sweet spot" where all the power is available. Most electric cars don't even have gearboxes.

So - how come gasoline engines for propeller-driven airplanes don't suffer the same problem? They don't even have shiftable gearboxes! Well, generally, they DO have variable pitch propellers - and that's a form of gearbox. It's actually a very efficient one because it has a continually adjustable "gear ratio" and no friction-creating gear teeth or clutch mechanism. This makes it much easier for the plane to keep it's motor running at the optimum RPM's.

Also, when acceleration is most important - on takeoff - the engine can be pushed to the optimum RPM, and the plane held on it's brakes until it needs to start moving. This means that the torque is already at the PERFECT amount when the plane starts rolling. With a car (and especially with a supercharger or a turbo) - the engine has to be at relatively low RPM's when you launch or you'll wreck the clutch or simply spin the wheels without much acceleration.

So many of the disadvantages of a gasoline engine aren't really a problem for airplanes - which makes the advantages of electric power very much less. For a car, the liability of a gasoline engine is a nightmare - and switching to electric motors solves a ton of issues.

Having said all of that - how much better is the acceleration of an electric car? Consider the 2013 BMW 1 series - which is the basis of the BMW ActiveE electric car. Let's consider the stats for a 2013 BMW 1 versus the Active E:

• Weight: BMW1: 1500kg, ActiveE: 1800kg...a 20% overhead for the electric version. Of course a tankful of gasoline adds another 50kg to the BMW1 - so it's a bit less than that.
• Torque: BMW1: 400Nm, ActiveE: 250Nm...oh...not good. But that's the PEAK torque for each car - and the ActiveE maintains that torque over a larger RPM range.
• 0-100kph time: BMW1: 6.3 seconds, ActiveE: 8.6 seconds - actually, better than you'd expect from the numbers given above. With 20% more weight, we'd expect 20% less acceleration for the same torque - but it also has vastly less peak torque. If BMW had put more powerful motors into the ActiveE, it could easily beat the BMW1 in a 0-100kph drag race - but with that much more weight, it's all-important recharge range would be much worse.

So read what you like into that.

SteveBaker (talk) 17:26, 11 April 2016 (UTC)

So the 0 to 60mph time for a Tesla is almost identical to it's 60 to 120mph time. -- I very much doubt that. With constant torque, you can bring constant force onto the road. But you need not double, but 4 times the kinetic energy when you go from 60 mph to 120 mph. --Stephan Schulz (talk) 17:53, 11 April 2016 (UTC)

You may doubt it, but it's correct. Assuming the torque remained constant, which it can do with an electric motor, then the force pushing the car forward is constant. F=ma, therefore the acceleration is constant. v=u+at: therefore constant acceleration gives identical change in velocity in a given time. The power is irrelevant to this: what actually happens is that in order to deliver constant torque, the motor mush deliver increasing power with rotational velocity.--Phil Holmes (talk) 08:31, 12 April 2016 (UTC)

might be true in a vaccum or something...but drag would indeed quadruple, I think...so the car would have less torque available for acceleration and more being used to maintain speed.....68.48.241.158 (talk) 18:14, 11 April 2016 (UTC)

Here's someone's graph of torque vs speed for a Model S. The motor is torque-limited (with power rising linearly) up to a certain speed, then power-limited (with torque dropping inverse-linearly). The crossover point is around 60 km/h ≈ 40 mph. So the Model S's 0–20 time should be similar to its 20–40 time (in a vacuum, at least), but 60–120 is a different matter. -- BenRG (talk) 21:14, 11 April 2016 (UTC)

Thanks to everyone for the responses. In short, batteries are the weakest link for the poor thrust to weight ratios? If we had some kind of super duper capacitor that had the energy density of gasoline, all our thrust to weight issues would be solved? ScienceApe (talk) 01:29, 13 April 2016 (UTC)

Related question, is it possible to have some kind of electric/jet fuel hybrid jet engine where you have an electric motor(s) spinning the turbines and ignite jet fuel to heat the compressed air? ScienceApe (talk) 01:33, 13 April 2016 (UTC)