Content deleted Content added

Inline

Revision as of 02:22, 29 June 2017

Welcome to the science section
of the Wikipedia reference desk.

skip to bottom

Select a section:

Shortcut

WP:RD/S

Want a faster answer?

Main page: Help searching Wikipedia

How can I get my question answered?

Select the section of the desk that best fits the general topic of your question (see the navigation column to the right).
Post your question to only one section, providing a short header that gives the topic of your question.
Type '~~~~' (that is, four tilde characters) at the end – this signs and dates your contribution so we know who wrote what and when.
Don't post personal contact information – it will be removed. Any answers will be provided here.
Please be as specific as possible, and include all relevant context – the usefulness of answers may depend on the context.
Note:
- We don't answer (and may remove) questions that require medical diagnosis or legal advice.
- We don't answer requests for opinions, predictions or debate.
- We don't do your homework for you, though we'll help you past the stuck point.
- We don't conduct original research or provide a free source of ideas, but we'll help you find information you need.

Ready? Ask a new question!

How do I answer a question?

Main page: Wikipedia:Reference desk/Guidelines

The best answers address the question directly, and back up facts with wikilinks and links to sources. Do not edit others' comments and do not give any medical or legal advice.

June 25

Species in which females are aggressive and males passive

Are there any species in which females are aggressive while males are passive?Uncle dan is home (talk) 04:49, 25 June 2017 (UTC)[reply]

The spotted hyena is a species of hyena that is very unique in the class Mammalia in that the females of the species have fewer nipples , lack an external vaginal opening, and possess a pseudo-penis. (Essentially they possess a very large clit; keep in mind the anatomical and development similarities between the clitoris and the penis.) The species operates in a matriarchal social system; the females are larger than the males and dominate over them. Even the lowest ranking females of the group have dominion over the highest ranking males. The female spotted hyenas tend to stick together over a large period of time while the male does not often form strong social ties to the den and is more migratory. A mother hyena is extremely protective of her cubs especially towards male hyenas. Penile erection is a sign of submission the male displays to a female hyena. Additionally, the spotted hyena is considered to be one of the most intelligent species of mammal in the world, ranking as smart or smarter than many species of primate. UltravioletAlien (talk) 06:18, 25 June 2017 (UTC)[reply]

Homo sapiens? Greglocock (talk) 06:45, 25 June 2017 (UTC)[reply]

Many species of hymenoptera. In bees, for example, the males (drones) only exist to mate, then they die. The female queen and workers run the place. StuRat (talk) 14:36, 25 June 2017 (UTC)[reply]

Popular culture is well acquainted with the black widow; among insects I think the preying mantis also. But the anglerfish is stranger than them all. Wnt (talk) 16:36, 25 June 2017 (UTC)[reply]

Aren't there some ~~grasshoppers~~ where the female decapitates the male after mating ? "Would you care to come in for a decap... uh, nightcap, I mean." StuRat (talk) 05:40, 26 June 2017 (UTC)[reply]

Actually Preying mantids, although subsequent research showed that this happens far more frequently when they're in captivity than in the wild, as the article discusses. {The poster formerly known as 87.81.230.195} 2.221.82.167 (talk) 11:44, 26 June 2017 (UTC)[reply]

Thanks, that's what I was thinking of. StuRat (talk) 05:43, 27 June 2017 (UTC)[reply]

Physics: Can a changing spacetime be used to generate unlimited energy?

I discovered that energy is not conserved in general relativity after reading this by Sean Carroll. He seems to argue that energy is not conserved if spacetime is changing.

The work done on a mass by a force F over a distance L is

$\int _{0}^{L}F\cdot dx$

If energy is not conserved in general relativity due to a changing spacetime, I wondered if it could be exploited to generate unlimited energy. The first example I considered was Hubble's law. Hubble's law says

${\dot {x}}=Hx$ .

Taking the derivative of both sides gives

${\ddot {x}}=H{\dot {x}}=H^{2}x$ .

Now, by Newton's second law and special relativity

$F={\frac {m}{\sqrt {1-{\frac {{\dot {x}}^{2}}{c^{2}}}}}}{\ddot {x}}$ .

But work can only be recovered up to the Hubble radius, so

$L={\frac {c}{H}}$

Substituting all of that and evaluating the integral gives

$\int _{0}^{\frac {c}{H}}{\frac {m}{\sqrt {1-{\frac {(Hx)^{2}}{c^{2}}}}}}H^{2}x\cdot dx=mc^{2}$

according to WolframAloha.

The energy generated is exactly the energy content of the mass used to generate that energy!

In fact, if you use different rules for the expansion of space (e.g. ${\dot {x}}=Hx^{2}$ ), you get the same answer ( $mc^{2}$ ). Here's the WolframAlpha for the example ${\dot {x}}=Hx^{2}$ .

This means you can't really use the expansion of space to generate energy, as the energy content of the mass used to generate that energy is the same as the amount of energy generated (and that mass disappears over the horizon once it reaches c/H).

That made me wonder if mass really IS conserved in a sense in general relativity, or if there is some other law preventing the generation of unlimited energy.

I know only the basics of physics. If you are an expert on physics (or if you know someone who is an expert on physics), can you let me know if a changing spacetime can or can't be used to generate unlimited energy? Perhaps a generalization of my reasoning above?

PeterPresent (talk) 09:41, 25 June 2017 (UTC)[reply]

The blog piece you quote seems very misleading. I mean, directly contrary to what he says, there is a difference between saying that energy is not conserved, versus saying that energy is conserved when the negative energy of the gravity field is taken into account. The difference is that you understand right off with the second option that if you make any circular set of changes to the gravity field, if in the end you have the same field you started with, you have the same energy you started with.

But the key issue here is what he says near the end:

First, unlike with ordinary matter fields, there is no such thing as the density of gravitational energy. The thing you would like to define as the energy associated with the curvature of spacetime is not uniquely defined at every point in space. So the best you can rigorously do is define the energy of the whole universe all at once, rather than talking about the energy of each separate piece. (You can sometimes talk approximately about the energy of different pieces, by imagining that they are isolated from the rest of the universe.) Even if you can define such a quantity, it’s much less useful than the notion of energy we have for matter fields.

If that is true, then it certainly seems like a worthy challenge for theoreticians. Surely there must be some way to divide up this energy neatly, without making any simplifying assumptions ... mustn't there? And if you have that math worked out, it seems like there ought to be other uses for it. Wnt (talk) 16:49, 25 June 2017 (UTC)[reply]

law of conservation of energy only applies to isolated system. A system in a changing gravity field (i.e., a changing spacetime, in relativity framework) is NOT isolated, and can gain (or lose) energy (for instance : pendulum energy is mgh, if g increases, so does energy). No big deal, even in a newtonian framework.
This law of conservation itself derives from the principle of least action, that general relativity also respects (ie: Einstein had not to get rid of this principle, he only had to adopt Einstein–Hilbert action instead of previous action (physics)).

So you wont generate energy through changing spacetime: you need as much energy to change space-time as it would get through the change. No free lunch (according to current theory, that seems to work pretty well, so we have no reason to change).

Gem fr (talk) 14:21, 26 June 2017 (UTC)[reply]

Feynman Lectures. Exercises PDF. Exercise 7-4 JPG

copyvio, see talk

The following discussion has been closed. Please do not modify it.

. .

...
DELETED We now turn to the case of elliptic orbits. This is essentially about three ellipses: both bodies move along ellipses (light-heavy, heavy-on-small) and, in addition, the relative motion of bodies also occurs along an ellipse (see Lectures, Chapter 7 ). All three ellipses are similar to each other, that is, they share the same eccentricity. If we also take into account that the center of mass of the system remains stationary (it lies in the common focus of the orbits of both bodies), and the distances from the center of mass to the bodies are inversely proportional to their masses, we come to the conclusion that the arrangement of bodies and their orbits is as in figure.

Denoting by v_1 and v_2 the velocities of the bodies M_1 and M_2 at the time when they are at apogee. As can be seen from the figure,

${\tfrac {v_{1}}{v_{2}}}={\tfrac {a_{1}+c_{1}}{a_{2}+c_{2}}}={\tfrac {a_{1}(1+e)}{a_{2}(1+e)}}={\tfrac {a_{1}}{a_{2}}}$
(Indices 1 and 2 denote quantities related to ellipses along which the masses M_1 and M_2 move).

To get the same expressions for elliptical orbits as for circular ones, remember that an ellipse can be imagined as a circle visible at some angle to its plane or (which is the same thing) as the projection of a circle onto an inclined plane. In other words, an ellipse can be obtained from a circle if you change the scale along one of the coordinate axes. Acceleration of the body while moving it along the circumference was calculated in "Lectures" (fig. 7.4). To obtain the acceleration of the body (for example, M_1) in the case of interest to us, imagine that its orbit is obtained from a circular increase in the scale in the "vertical direction" a_1 / b_1 times. The value of x will not change, and s will increase and become equal to s₁ = (a₁ / b₁) s.
Substituting x² = 2Rs (valid for the circle) their values after increasing the scale x₁ = x, s₁ = (a₁ / b₁) s and R = b₁ (The "horizontal" dimensions have not changed, so the small semiaxis of the ellipse is equal to the radius of the original circle), we obtain

$x^{2}=2{\tfrac {b_{1}^{2}}{a_{1}}}s_{1}$

Thus, the radius of curvature of the ellipse at the point of intersection with the semimajor axis is ${\tfrac {b_{1}^{2}}{a_{1}}}$ . Assuming that for a very small period of time the first body moves along a circular orbit of this radius, we can write

${\tfrac {v_{1}^{2}a_{1}}{b_{1}^{2}}}={\tfrac {GM_{2}}{(a+c)^{2}}}={\tfrac {GM_{2}}{a^{2}(1+e)^{2}}}$

(Here a and c are the orbital parameters of the relative motion of the bodies: a = a_1 + a_2, c = c_1 + c_2). Similarly for the second body

${\tfrac {v_{2}^{2}a_{2}}{b_{2}^{2}}}={\tfrac {GM_{1}}{a^{2}(1+e)^{2}}}$

Adding the last two equalities and expressing v_2 in terms of v_1, we obtain

${\tfrac {v_{1}^{2}(1+e)}{a_{1}^{2}(1-e)}}={\tfrac {G(M_{1}+M_{2})}{a^{3}}}$
It remains only to find out what relation the quantity on the left-hand side of this equation has to the period of revolution. First of all, note that the area that the radius vector of the body M_1 (drawn from point 0) "sweeps out" per unit time is 1/2 v_1 (a_1 + c_1) = 1/2 v_1a_1 (1 + e). Although in fact we calculated the rate of change of the "swept out" area for the moment when the body M1 is at apogee, according to Kepler's second law, this speed does not change when the body moves in orbit. Therefore, the value 1/2 v_1a_1 (1 + e) T (here T is the period of revolution) is equal to the area of the orbit of the body M_1. The area of the ellipse is easy to calculate if you realize that when you zoom in on one of the axes, the area of the figure increases by the same factor as the scale. Therefore, the area of the ellipse is
$\pi b_{1}^{2}{\tfrac {a_{1}}{b_{1}}}=\pi a_{1}b_{1}=\pi a_{1}^{2}{\sqrt {1-e^{2}}}$

It is now easy to see that
$T={\tfrac {2\pi a_{1}}{v_{1}}}{\sqrt {\tfrac {1-e}{1+e}}}$ , and $T^{2}={\tfrac {4\pi ^{2}a^{3}}{G(M_{1}+M_{2})}}$

— MEPhI , Solutions (Google Translate)

Apparent WP:COPYVIO, see talk μηδείς (talk) 23:27, 25 June 2017 (UTC)[reply]

Here is my solution png. The question is can we use 3rd Kepler's law for relative circular orbit and relative elliptical orbit? Or the 3rd Kepler's law is just approximation.

Username160611000000 (talk) 12:46, 25 June 2017 (UTC)[reply]

See Kepler's third law -- it is an approximation. Wnt (talk) 16:53, 25 June 2017 (UTC)[reply]

Hm, they uses 2nd Kepler's law during the proof. So we cannot derive the exact formula for circular orbit and then extrapolate the approximation formula on case of elliptical orbits, can we? Username160611000000 (talk) 09:46, 26 June 2017 (UTC)[reply]

The second law should not be an approximation so long as the right center is used - the actual barycenter in the case of a two-body problem, rather than the center of the larger body. This is because of conservation of angular momentum. Naturally, any torque from additional masses would tend to throw it off. Wnt (talk) 17:12, 26 June 2017 (UTC)[reply]

Looking at the problem myself, you have a force of GmM/D^2 pulling inward. Note I use D for the distance between them which is only approximately r (the radius of the circle) for the smaller mass. The centripetal acceleration needed is -w^2*r, where w is angular velocity (I'll be lazy and avoid the omega). The second is a vector; the first is a magnitude, so I'll wave my hands and make the "-" go away, then convert acceleration to force by multiplying by the mass, leaving me with GmM/D^2 = mw^2*r for the little mass = and GmM/D^2 = Mw^2*r for the big one. But what is r? Well, it's a (literally) weighted average of the full distance D for one mass and 0 for the other, i.e. (0*m + D*M)/(m+M) or (D*m + 0*M)/(m+M). The first (DM/m+M) defines r=0 at the small mass and the second (Dm/m+M) puts r=0 at the large. Either way we flip that and get GmM/D^2 = mw^2*DM/(m+M) = Mw^2*Dm/(m+M), depending on which mass you consider. So G(m+M)/D^3 = w^2. The period is 2 pi w, so this is a period inversely proportional to D^(3/2) and directly proportional to the square root of m+M.

Note that we use the inaccurate Kepler's third law version in the first section of orbital period, without warning people it is wrong, even saying that mu depends on the 'more massive body' rather than 'total mass'... then get out the right version in the third section down. I think I actually tried to fix that once and got rebuffed, but there is an article that looks way more mysterious than it ought to if anyone wants to hack at it. Wnt (talk) 17:44, 26 June 2017 (UTC)[reply]

Bringing home the process of making industrial nutritional yeast - how to do it?

I know very little about the industrial process of making nutritional yeast other than the reading on Wikipedia. Some kind of unnamed strain is used to culture the yeast cells on top of the agar plate, I presume. The agar plate probably contains glucose from molasses or sugarcane, which is food for the nutritional yeast. Then, the yeast cells are heat-deactivated, harvested, washed, dried, packaged, and sold in grocery stores. In a home kitchen, an agar plate can probably be replaced by a tomato sauce jar lid. Maybe the agar can be replaced by gelatin (may be easier to buy than agar). The person may want to heat-sterilize whatever metal (has to be metal, as plastic will just melt) instrument he's using and layer the Baker's yeast (same species, but may be a different strain) on the culture plate with the sterilized rod. The culture plate (lid) may be screwed back on the tomato sauce jar container, but the container is placed upside-down, because of the yeast cells. Incubate (in the conventional oven?) for 24 hours or until the culture plate seems to be covered by yeast. Then, the person should select for "desirable" phenotypic characteristics. But, what is considered desirable? Maybe it's best to harvest everything? For the washing process, a cheesecloth, perhaps, may be used to filter the yeast, but that assumes that the yeast will not fall through the holes of the cheesecloth. For drying, the yeast cells may be gathered into a jar. The jar is mostly sealed except for a space to allow a blow-dryer in and dry the yeast cells. Afterwards, the yeast may be used as topping on foods.

I have a number of concerns. (1) Does the strain of yeast really matter? Can Baker's yeast strain replace the Nutritional yeast strain? (2) Is the umami flavor of Nutritional yeast caused by the glutamate release of the cells? (3) Will the protocol above actually work as planned? Are there any holes or false assumptions (such as the cheesecloth which may have holes that are too large for the yeast cells)? (4) What "desirable" characteristics are people selecting for when they make nutritional yeast on an industrial scale? 50.4.236.254 (talk) 19:09, 25 June 2017 (UTC)[reply]

As a side question, what is the formal job title of the person who cultures the plates industrially? How can one apply to such a position at a company? 50.4.236.254 (talk) 19:21, 25 June 2017 (UTC)[reply]

I'm sure, without even looking, that industrial yeast is raised in a broth and filtered out. I'll take a look a bit later. Abductive (reasoning) 21:27, 25 June 2017 (UTC)[reply]

As you know there are different strains of yeast. So the starting culture has its growing medium, tailored to suit the type of yeast that one wants to be in abundance ( by adjusting the mediums pH and things), (which insistently, did not require modern science – just experience gained over many generations). By this process, one can produce starting cultures selected for either bread making, wine, beer, larger, etc. This was achieved long before the invention of the microscope. Aspro (talk) 23:34, 25 June 2017 (UTC)[reply]

Just so you know, I don't think that's an answer to my question. My question is not about the different strains of yeast. I already know that. Nor is it about the process of making different strains of yeast in a pre-modern fashion. It is, at the most basic level, about whether deactivated baker's yeast/brewer's yeast/wild yeast can replace "nutritional" yeast and still have that umami-rich flavor. 50.4.236.254 (talk) 11:27, 26 June 2017 (UTC)[reply]

[Adding the 'close small' coding that 50.4.236.254 forgot, thus rendering the entirety of the following queries small. {The poster formerly known as 87.81.230.195} 2.221.82.167 (talk) 11:39, 26 June 2017 (UTC)][reply]

A liquid growing medium should be far more productive that a plate. Grow the yeast in a jar full of a sugar solution with added micronutrients. Adapt methods used for making ginger beer to optimize yeast growth. The yeast will settle in the bottom as brown "sludge". Roger (Dodger67) (talk) 08:12, 27 June 2017 (UTC)[reply]

June 26

A guy where I work contends that CSPAN is actually harming democracy and functioning of usa legislature.

He says because of cspan, politicians can't make deals, and can't have personal friendships across party lines. Is this true? Also, are there political scientists and journalists who have discussed this claim, either pro or con?65.103.249.243 (talk) 00:00, 26 June 2017 (UTC)[reply]

In the wake of the shooting of a Republican congressman, I heard some members of both parties talk about how they all get along personally, they just differ over policies. ←Baseball Bugs ^{What's up, Doc?} carrots→ 00:33, 26 June 2017 (UTC)[reply]

Question transferred to the Humanities Desk, whose topics include politics—here. --76.71.5.114 (talk) 00:33, 26 June 2017 (UTC)[reply]

Engine knocking

According to the engine knocking article, the phenomenon is distinct from, but routinely follows, instances of pre-ignition. Lower in the article, I read that "pre-ignition can destroy an engine in just a few strokes of the piston". Presumably you wouldn't have knocking if the engine were destroyed, so how does this work? Is it simply that pre-ignition sometimes doesn't destroy the engine in such a short period of time, so the knocking has a little time to manifest before the whole thing becomes useless? Nyttend (talk) 00:04, 26 June 2017 (UTC)[reply]

I think you are reading too much into it. I know of one engine that runs in preignition and knock for 10s of hours on the dyno, as a durability test. I suspect a liberal sprinkling of ^{[citation needed]} tags in that article would reveal that it is mostly factoids from on line forums. Greglocock (talk) 03:39, 26 June 2017 (UTC)[reply]

It depends on how far advanced in the power cycle the knocking or pre-ignition is taking place. The effect of a slightly advanced (a few degrees of rotation) spark can usually be heard when the accelerator is pressed hard under load. This causes engine wear, because the spark takes place while the piston is still rising on its compression stroke, and rings, pistons and big end bearings are subjected to loads they were not designed for. Obviously, the more advanced the spark, or the more advanced the pre-ignition due to other causes, the greater will be the damage. It's not unusual for piston rings to be cracked as a result. Sometimes, small broken pieces of the rings manage to fall out of their grooves and if they get sucked up by the oil pump, they may ruin the pump's gears very quickly or travel through the oil bores and score crankshaft or camshaft bearings. A kind of compounding of the problem may develop: the more advanced the spark, the weaker the performance of the engine, leading the driver to put his foot down harder on the accelerator, and causing more wear. Akld guy (talk) 05:42, 26 June 2017 (UTC)[reply]

Knocking is after spark plug ignition, pre-ignition (also called rumble) is before spark. The basic reason for either is if the gas-air mixture stays too hot for too long it auto-ignites (depending on thermodynamical parameters), which is usually not desired as the spark allows combustion control. Knock autoignition is easier, because as the flame front develops around the spark plug gases outside the flame get compressed and heated a bit, plus there is more time for the auto-ignition delay to run out. Obviously, if you can have the hard thing (pre-ignition) you can have the easy thing (knock) unless pre-ignition is triggered by a solid surface that is a lot hotter than other places in the engine (e.g. the exhaust valve), so knock often follows pre-ignition.

The problem in both cases is that the pressure buildup in the cylinder takes place faster than intended, which can damage the engine. Pre-ignition is worse because it makes more gas burn in the uncontrolled mode and hence the pressure buildup is faster. Some modern engines are intended to run in permanent "preignition" mode but most are not (and consequently the dimensions, material strengths etc. cannot withstand rumble or prolonged knock). Tigraan^{Click here to contact me} 11:01, 26 June 2017 (UTC)[reply]

It's mostly a question of BMEP. The worse the engine is for predilection to knocking, the less damaging it can be. If the engine knocks at low loads, then the forces arre low and nothing gets damaged. If an engine is heavily loaded though, without knocking, and then suddenly begins to knock (such as a fault in the timing suddenly occurring, or failure in something like a water injection system), then it can indeed fail in a few strokes. Many engineers will have started out with a single cylinder variable compression research engine, built like a brick cowshed, in a test cell at university, happily listening to it knocking all day whilst demonstrating high compression and low fuel octane. They'll do it indefinitely, that's what they're for. But tune the 'nads off something highly strung, have some part of the control gear lose the plot momentarily, and it can be expensive.

That's a dreadful article though, hopelessly confused. To understand the fundamentals, there's still no substitute for sticking down quietly with the old editions of Ricardo and reading them cover to cover. Andy Dingley (talk) 11:18, 26 June 2017 (UTC)[reply]

Do you mean these books by Harry Ricardo 1885 - 1974 ? Blooteuth (talk) 15:05, 27 June 2017 (UTC)[reply]

Yes, particularly the 2nd, 3rd and 4th editions. They're not simply editions, they're pretty much different books (and a different author for the 3rd). The pre-war editions are expensive, late copies (and the modern paperback) are OK. Andy Dingley (talk) 15:28, 27 June 2017 (UTC)[reply]

Engine knocking on gas engines causes damages. Diesel engines rely on the self ignition. Diesel fuel burns slower. A cold diesel engine is hard and dirty to start. For this reason, have the glowplugs heat up, before turning the starter. Engine knocking on gas engines occurs form early ignition or wrong engine timing. Incomplete burned fuel or engine oil which came into the cylinder head by any other failure is still glowing from previous ignition. This ignites the new fuel earlier than the controlled ignition over the spark plug. This effect can be seen like stepping the pedal of a bike before it reaches the top position. As the engine has no free run, the pressure of the combustion forces a little reverse to the drive shaft but most against the curb shaft bearings and the cylinder head. Increasing the thermodynamic efficiency of the engine needs to use a more lean gas-air mixture, causing more heat drop on the cylinder and cylinder head. This is the reason for installing a knocking sensor into the engine. When knocking is detected, the mixture ratio will be changed by the engine controller unit. --Hans Haase (有问题吗) 17:54, 27 June 2017 (UTC)[reply]

Knocking is not the ignition process in a diesel engine. Nor is ignition in a diesel engine, or even a semi-diesel engine, the same process as knocking in a petrol engine. Andy Dingley (talk) 18:06, 27 June 2017 (UTC)[reply]

See THIS comment!

{{xt|Would not the brightness at higher latitudes and elevations be affected by snow and ice? In the Arctic the glare can be painful. 15:05, 26 June 2017 (UTC)

The discussion is at Wikipedia:Reference desk/Archives/Miscellaneous/2017 February 3#Daylight intensity by month?. 92.62.8.1 (talk) 15:05, 26 June 2017 (UTC)[reply]

See snow blindness Wymspen (talk) 15:17, 26 June 2017 (UTC)[reply]

Feynman Lectures. Exercises PDF. Exercise 7-11 JPG

. .

...
7-11. In making laboratory measurements of g, how precise does one have to be to detect diurnal variations in g due to the moon's gravitation? For simplicity, assume that you laboratory is so located that the moon passes through zenith and nadir. Also, neglect earth-tide effect.

— R. B. Leighton , Feynman Lectures on Physics. Exercises

First, I can't understand how the laboratory can be just below and above moon (zenith and nadir). It is only possible if the north and south poles of the earth are equidistant from the moon (e.g. at vernal equinox moon is on the line sun-earth) and simultaneously the laboratory is situated on the earth equator. Does Feynman mean this?

To solve the problem we should say how many decimal digits we need. And it returns us to my unanswered question here.

Now I understood what you meant by the word "lag". There is a lag 6 hours between the bulge and the moon-earth line. But first I must understand the bulge appearance (and height) explanation and only then consider the lag. E.g. I don't understand why is the tides effect explained by the gap of (4.51 - 4.36)×10^-5 but much bigger gap of equator point is ignored. The point on equator moves through lunar orbit plane. When the point intersects the plane, acceleration = 4.36×10^-5 , but at leftmost and rightmost positions acceleration = 3.38 ×10^-5 [1]. So gap =(4.36 - 3.38)×10^-5 . Username160611000000 (talk) 20:21, 13 June 2017 (UTC)[reply]

Username160611000000 (talk) 16:47, 26 June 2017 (UTC)[reply]

The exercise is intended to measure the difference in gravity felt on Earth as the Moon gets closer and further away. The Moon is not always exactly the same distance from the Earth. It get closer and further away, which is the zenith and nadir. So, for simplicity, we assume that your lab is directly below the Moon when it is at the zenith and directly below the Moon when it is at the nadir. In other words, you will measure the absolute maximum difference in gravity change because you will be at the point where you have the maximum difference between close and far distance measurements. 209.149.113.5 (talk) 17:08, 26 June 2017 (UTC)[reply]

According to [2] the lunar tidal acceleration at the Earth's surface along the Moon-Earth axis is about 1.1 × 10⁻⁷ g. If we calculate g_max/g_min = (384 400 + 6400)² / (384 400 - 6400)², we get 1,068 or difference = 6.8× 10^-2g what is absolutely distinct from article value 1.1 × 10⁻⁷ g. Username160611000000 (talk) 17:26, 26 June 2017 (UTC)[reply]

Hehe, which g is that? You just calculated how much the gravity of the moon varies, but now you want to compare it to the gravity of the Earth. Though I'll admit I didn't look up what the ratio of those two is to make sure this works out right. Wnt (talk) 19:28, 26 June 2017 (UTC)[reply]

Thank you. It was mistake.
g_max = 6.67•10^-11 • 7.35• 10²² / ((384 400 - 6400)• 10³)² = 3.43• 10^-5 m/sec²
g_min = 6.67•10^-11 • 7.35• 10²² / ((384 400 + 6400)• 10³)² = 3.21 • 10^-5 m/sec².Username160611000000 (talk) 20:17, 26 June 2017 (UTC)[reply]

Difference = 3.43• 10^-5 - 3.21 • 10^-5 = 2.2 • 10^-6 m/sec² = 2.2 • 10^-7 • 9.8 m/sec² = 2.2 • 10^-7 • g. It seems, article uses this non-uniform field , generated by the moon. But it's wrong. On the surface of the earth there is a centrifugal acceleration due to revolution about the barycenter and a gravimeter counts it. Username160611000000 (talk) 12:17, 27 June 2017 (UTC)[reply]

There's no mistake. The centrifugal acceleration does not show the diurnal ("daily") variation that the question asks about. R. B. Leighton did not mispell "you[sic] laboratory". Blooteuth (talk) 14:52, 27 June 2017 (UTC)[reply]

The earth is rotating about its axis, so the centrifugal force applied to a piece of the earth (e.g. 1 kg of rock) will not change its direction and magnitude (period = 27,3 days) , but earth surface will move (period = 1 day). Therefore there are two fields : one due to gravitation, second due to centrifugal force. Both fields have a period of 27 days. There is no difference in the effects shown by these fields. Username160611000000 (talk) 16:22, 27 June 2017 (UTC)[reply]

Now you've got me confused. ;) An object orbiting about a barycenter generally does not experience centrifugal force (in its rotating frame of reference), because that force is counteracted by gravity. The whole point of an orbit is to keep the bodies apart by "centrifugal force", you might say. On the other hand, it does occur to me that a phase locked body like the Moon does move its far side further each revolution than its near side. So ... at least by the Kepler third law approximation ... the far side is in the "wrong orbit", whereas the center of the Moon is just right. Well, I think this greater "centrifugal force" is just the flip side of the tidal gradient seen in a usual calculation made in a non-rotating frame, but now I'm not entirely sure. Wnt (talk) 16:01, 27 June 2017 (UTC)[reply]

An object orbiting about a barycenter generally does not experience centrifugal force (in its rotating frame of reference), because that force is counteracted by gravity. It is correct for material point, and as Feynman mentions moon’s attraction just balances the “centrifugal force” at the center of the earth [3]. We should return to the previous discussion and consider a model in which the earth is made up of 1 kg bricks. The brick on the far side of the earth feels gravitation force almost the same as at the center (lower by 1.1 • 10^-6). To stay on the circular orbit the brick needs centripetal force ω²R , since R is bigger almost 2 times on the far side, ω must be lower 1.5 times. But the earth is a rigid body, so rotates about the barycenter with averaged ω (sum over all bricks of centripetal forces must be equal to the sum over all bricks of gravitational forces). Therefore the earth pulls the far-side brick. If the molecular or gravitational forces acting between the bricks had disappeared for a time, then the brick would have flown away upward. Username160611000000 (talk) 16:22, 27 June 2017 (UTC)[reply]

Have there ever been lunar calendars where a solar eclipse makes the month start early?

Presumably they would have to not set before the moment of New Moon else that'd just be the old Moon. Sagittarian Milky Way (talk) 22:52, 26 June 2017 (UTC)[reply]

Have there ever been any calendars where eclipses have any function whatsoever? --jpgordon^{𝄢𝄆 𝄐𝄇} 02:33, 27 June 2017 (UTC)[reply]

I don't know but lunar calendars start when the Moon is seen or calculated to be seen. The world record for young crescent sighting is about 11.5 to 15.5 hours after New Moon (depending on "found with naked eye", "binoculars to find it", "binoculars to see it", "telescope", "high altitude, computer to aim telescope just to find it, have to hide under a sheet for x minutes first or you're not dark adapted enough", that kind of thing). However you can see a c. 0-1 hour old Moon during a solar eclipse. Sagittarian Milky Way (talk) 03:19, 27 June 2017 (UTC)[reply]

Solar eclipses always occur at new moon. In the Hebrew calendar, at least, new moon marks the beginning of the month (they no longer use the crescent moon method). ←Baseball Bugs ^{What's up, Doc?} carrots→ 03:03, 27 June 2017 (UTC)[reply]

There are limitations on the length of the lunar month. It may not have fewer than 29 days or more than thirty. So I doubt that a solar eclipse would prompt anyone to declare a new month beginning at nightfall if it happened on the 28th day. There might be something in Babylonian records, but in our culture, when this happened the actual date of the eclipse was suppressed Computus#History. If the regulation is impeded by clouds the new month will start at the end of the thirtieth day. There could be a succession of months where the starting date was empirically determined in this way - if a new moon appeared at the close of the 28th day I wonder how the regulators would handle the situation. 94.195.147.35 (talk) 09:15, 27 June 2017 (UTC)[reply]

According to this a solar eclipse cannot occur at the beginning of Ramadan, because it takes a day or two for the crescent to become visible. So at least that lunar calendar rules it out, even though it relies on direct observation of a crescent. (Then again, there are often schisms in Islam, and I've noticed that sources about it seem uncommonly willing to omit the differing point of view). I haven't heard of another lunar calendar where people were actually watching to sight a fresh crescent, but I don't know them all! Wnt (talk) 16:06, 27 June 2017 (UTC)[reply]

That's how the lunar and lunisolar calendars all started in the past though, actual sighting. Though this issue wouldn't turn up before a culture realized a solar eclipse is the Moon and thought it was a dragon eating the Sun or something. Sagittarian Milky Way (talk) 19:37, 27 June 2017 (UTC)[reply]

If you were trying to ask whether all Muslims use direct observation, the answer is no. Our article isn't well sourced but does cover this Islamic calendar#Astronomical 12-moon calendars although as also mentioned there, not all of them use it for religious purposes. See also Tabular Islamic calendar and [4]. Both Islamic calendar#Astronomical considerations and Islamic calendar#Theological considerations also cover the various controversies and debates. As mentioned there and covered by this French source [5] some of the observations or decisions seem questionable based on what is known and what the country claims to do. However the majority of Muslims do theoretically follow the observation method at least for religious purposes. Nil Einne (talk) 11:15, 28 June 2017 (UTC)[reply]

June 27

Can anyone have in their mind a representation of four or more dimensions?

Can people (who presumably study higher dimensionality) hold in their mind a representation of those dimensions like I think I can just about hold in my mind three-dimension arrangements? ----Seans Potato Business 17:07, 27 June 2017 (UTC)[reply]

It depends on what the meaning of "like" is. We don't have experience with four spatial dimensions, and I'm not aware of anyone who can visualize four-dimensional objects in the same way that you would visualize three.

But there are workarounds. Take a Klein bottle, for example. It can't be embedded into three-dimensional space without self-intersection. But you can, say, think of most of it as green, and the part that comes around to self-intersect shading into red, and then back to green, whereas the part it pokes through is green. Or you can imagine coming up to the self-intersection and then looking at the area rotated, so that you see a path passing around rather than through. Or you can map one dimension to time and think of an animation.

Five is obviously even harder. --Trovatore (talk) 17:41, 27 June 2017 (UTC)[reply]

Six is right out. Rojomoke (talk) 22:32, 27 June 2017 (UTC) [reply]

Some people claim to be able to do so, and that you can get better with practice. Charles Howard Hinton invented a set of colored blocks that he claimed could be used to train yourself to visualize the fourth dimension. See also: [6] CodeTalker (talk) 18:07, 27 June 2017 (UTC)[reply]

I either use time as my fourth dimension and have a four dimensional concept be a short movie, use color, or in the case of polytopes where the majority of vertices are in a limited number of planes, just offset them with dashed connectors. (So w+1 is equivalent to x+50).Naraht (talk) 19:18, 27 June 2017 (UTC)[reply]

Yes, using virtual reality glasses and practice. This is the second time this year this has been asked. μηδείς (talk) 19:55, 27 June 2017 (UTC)[reply]

@Medeis: I'm not sure what that link accomplishes, aside from helping Google track who reads this page. Some archives of relevance low high high low low. This is not a complete list because the Wikipedia search results give way too little context in the blurbs. The point of most of the "low" relevance ones I mentioned is that visualizing three dimensions is not actually all that simple. The eyes have a two dimensional surface, and the rest is tricks. These tricks are not really good enough to visualize an entire block of space unless you can make some assumptions (like it's a function with one value per 2D position). Nonetheless, tricks can also visualize 4D, under limited circumstances. Wnt (talk) 20:17, 27 June 2017 (UTC)[reply]

It is possible to be comfortable with tricks of the light, if that is all you have. — Hactar Wnt, that's nicely put. Although I would argue that we have a sense of three-dimensionality that goes beyond purely what can be represented in the retina. Still, intuition is to some extent extensible, and that's actually one of the great satisfactions of mathematics. --Trovatore (talk) 23:47, 27 June 2017 (UTC)[reply]

Of course our sense of volume and three-dimensionality is not purely due to retinal imaging. Blind people conceive of space too. There have been plenty of blind sculptors, architects, etc. Intuition is indeed extensible, and that's well put too. WP:OR For my research, I tend to wrangle fairly fixed sets of equations specifying dynamical systems, and explore 5-20 dimensional parameter spaces at a time, visualizing with various planar slices, color coding, surfaces in space, animations, etc. Put a few color-coded 3D animations in a stack and you get to 5D slices in a fairly straightforward manner. There's lots of room up there. And while I won't claim I can picture ~5D the same as 3D, I do build up intuition and make inferences that are supported by data when I get familiar enough with a certain system. I've tried many times over the years to find real empirical research on conceptualization methods of 4-n dimensional stuff but I always come up blank :-/ SemanticMantis (talk) 00:29, 28 June 2017 (UTC)[reply]

Depends on what 'anyone' means. IIF you mean can some people do it then the answer is yes. If you mean can everyone do it then the answer is no. Many people aren't even able to imagine 3D objects properly, a cube is about as easy as it gets but it isn't immediately obvious to them for instance what shape a corner of a cube stuck into plasticine would look like or that there's slices though a cube producing hexagons or trapezoids or five sided figures. Dmcq (talk) 22:23, 27 June 2017 (UTC)[reply]

4D is possible with effort, given the number of anecdotal reports that are easily found. Given that, 5D may be possible too. Beyond that point, there is always the joke about visualising 9D by visualising nD and setting n = 9. ^_^ Double sharp (talk) 23:46, 27 June 2017 (UTC)[reply]

Yes I don't know if anyone is able to imagine 5D well, even an image of it would be four dimensional. There's various tricks for adding extra dimensions to points like having a range of colors or time or other attributes and these can work fairly well if one doesn't want to rotate anything. Dmcq (talk) 10:38, 28 June 2017 (UTC)[reply]

I can sort of handle 4D, but it is difficult (though it gets easier with practice). I can't do 5D at all, though; I can mentally understand how it must go on, but the mental image gets more and more difficult due to the projections of projections necessary. I imagine that if you could grasp 4D with about the same level of sureness as we all grasp 3D, 5D ought to barely be within reach, but of course that is highly uncertain. Double sharp (talk) 11:00, 28 June 2017 (UTC)[reply]

For additional reference, there is a lot of scientific research on data visualization of high dimensional data. It is not exactly the same as a personally poking your mind's eye up in to N-dimensional clouds, but it is a bit easier to study. See e.g this [7] scholarly paper, this [8] google AI advertisement, and this [9] quora discussion. SemanticMantis (talk) 00:32, 28 June 2017 (UTC)[reply]

So, who are these people who can visualize how a left handed helix is going to become a right handed helix by rotating it along the axis orthogonal to it? Count Iblis (talk) 02:40, 28 June 2017 (UTC)[reply]

How is that much harder than the left no my left picture taking thing? I can't see how you can get a pentagon from cutting a cube by imagination alone. Sagittarian Milky Way (talk) 02:48, 28 June 2017 (UTC)[reply]

If you have a horizontal slice through a cube you get a square, If you pull up one corner of that plane so it goes past one corner of the cube you get a fifth side. It isn't a regular pentagon but it certainly is five sided. And yes you're right, a 3D helix is flat in the fourth dimension but it is still interesting to visualize. It is the difference between imagining a whole object and imagining what the object looks like from various points of view, I think of the first as feeling or internalizing and the second as seeing. Dmcq (talk) 10:11, 28 June 2017 (UTC)[reply]

The pentagon slice seems so obvious now lol. Sagittarian Milky Way (talk) 21:21, 28 June 2017 (UTC)[reply]

Hexagonal_crystal_family and Cubic crystal system will help you visualize how a plane cube slice can be a triangle, pentagon or hexagon Gem fr (talk) 13:53, 28 June 2017 (UTC)[reply]

Visual system is a must read here. Don't forget it turns 2-D images into 3-D perception, though a 2-D process.

I guess "hold in the mind a representation" means somethings like "being able to stimulates the visual cortex in a way similar to the way to would react to actually seeing a 4 or 5 (or more) dimension object". And of course it can be done, as far as mathematics are involved (and they are, pretty much):

think of fractal objects : you can fill a plane or a cube (and even a 4-D or n-D "cube" !) with a line , in such a way that any place in the plane/cube is very close to a particular point of the line. So that point of the line is a "representation" of that place of a plane/cube

The main obvious trouble is that the visual perception system is trained to "see" 3-D objects, neither 2-D nor 4 or 5-D objects. In my previous example, the point of the line will obviously be first interpreted as a point of a line, before a point in a plane, or a point in a cube.

I mean, any stimulation that could be associated with a 4-D object would ALSO be associated with a 3D or a 2D image, and those would naturally take precedence over the 4-D object representation. That's where training may interfere. Obviously upon seeing a flat 2-D image on a TV screen, we naturally interpret it as 3D, while it is not. So i guess, the very same way, trained people could "see" 4D objects when exposed to real or imagined 3D things.

Gem fr (talk) 12:55, 28 June 2017 (UTC)[reply]

There's quite a bit involved in seeing 3D. What is on the retina is like the 2D TV screen and yet we have to associate a distance with each point and get a feeling for a 3D world from it. We see a tree through a window and we know it is much farther away and bigger than the window even though its image is within that of the window. The same has to be done to see 4D. Dmcq (talk) 15:08, 28 June 2017 (UTC)[reply]

Limit of intellect

DNFT. See talk--Jayron 32 04:58, 28 June 2017 (UTC)[reply]

The following discussion has been closed. Please do not modify it.

Is the limit of the intellect be the limit of memory?--109.252.29.219 (talk) 20:08, 27 June 2017 (UTC)[reply]

As for me, the memory done mind.--109.252.29.219 (talk) 20:41, 27 June 2017 (UTC)[reply]

I would say no, intellect has little to do with memory. My reasons:

1) The ability to combine different bits of info together into useful concepts is how I would define intellect. Those "bits of info" don't need to be stored internally in our minds, at least since the invention of writing. Computers and the internet have greatly expanded our ability to access info outside our own brains.

2) Computers alone have vast memories, but that doesn't mean they have a vast intellect.

3) Some animals may have better memories than humans, such as whales and elephants. That would explain what they do with their much larger brains. However, again that doesn't give them a far superior intellect. I believe this is because they can't synthesize the bits of data at their disposal in the way humans can.

4) Of course, there is some minimal amount of memory required to be able to access the external info. For example, if you can't spell search terms well enough for Google to recognize what you are searching for, it won't find anything useful for you. StuRat (talk) 21:14, 27 June 2017 (UTC)[reply]

If there is a limit we haven't come anywhere near it yet. If at one end of the Bell curve a child of nine can pick up a university chemistry book and read it through and apply it - then what would be at the end of the Bell curve if that was just average? Dmcq (talk) 22:49, 27 June 2017 (UTC)[reply]

<Argentiniany music> He once won a rugby Test while replacing a star even though he's not athletic. He once beat Magnus Carlsen at chess. Blindfolded. He's memorized Shakespeare. He once bought the New York Yankees with short term capital gains. He invented the phaser. His jokes take a team of at least 5 diverse doctorates to understand. He is.. the most intelligent man in the world. I don't always get PhD's, but when I do, I prefer English unis. Stay percipient my friends. Sagittarian Milky Way (talk) 00:20, 28 June 2017 (UTC)[reply]

June 28

Characteristics of a predator

If a lone human gets eaten up by a crocodile, then is the crocodile the predator?
If a human kills a pig and eats its flesh, then is the human a predator?
If a human exchanges metal coins for a dead pig's flesh from a dead pig that is killed by another human or a slaughter machine, then is the human a predator? Or does being a predator automatically mean that the human must kill the animal and eat it?
If a very tiny population of creatures eat a human's flesh, then is the whole population of tiny things considered a "predator" or are they "many predators"? 50.4.236.254 (talk) 03:13, 28 June 2017 (UTC)[reply]

Small organisms that feed off of another living organism without killing that other organism are generally considered parasites rather than predators. Generally, a predator is an organism that obligately *kills* other animals to feed on them. Where this gets murky, of course, is in the case of parasitic animals that inevitably kill whatever they're parasitizing; these are called parasitoids (parasitoid wasps, for example). NorthBySouthBaranof (talk) 03:17, 28 June 2017 (UTC)[reply]

~~Agreed on the last point.~~(I retract that per Gem fr below; we both made unwarranted assumptions about the tiny things) For the others: technically, I think that exchanging metal coins for part of a dead pig makes you a scavenger, sure as if you had growled at the hyenas to drive them off. Of course in terms of digestive adaptations you may not look quite like a scavenger, but to some degree you might... human ancestors were notorious for running to where the buzzards were going and cracking bones for the marrow. Killing the pig is surely predation. And crocodiles... well, if a crocodile isn't a predator, what is? :) The term omnivore applies (humans aren't what you'd call carnivores or herbivores) but that doesn't rule out acting as the predator (just ask the grizzlies, they'll be happy to explain) Wnt (talk) 03:24, 28 June 2017 (UTC)[reply]

It should be noted that most predators are scavengers as well, which is to say that carnivores don't pass up a free meal even if they didn't kill it. This article notes that it's easier to thinking of carnivores as existing on a continuum of scavenger-predator and that there's few of any scavengers that never kill, nor predators that don't scavenge. --Jayron 32 05:15, 28 June 2017 (UTC)[reply]

Also as is nearly always the case, definitions are fussy at the boundaries. In particular, I don't think it's clear whether something that eats prey hunted in a community setting is a scavenger. Is a lion cub a scavenger when it eats food caught by others in the pride in a hunt it didn't participate in? What about a male lion in a pride? (As our article mentions Lion#Hunting and diet group hunts in a pride generally involve the lionesses although male pride lions probably hunt more than we're previously appreciated.) When a lion (or lion pride) steals the prey of a hyena sure. For that matter if it steals the prey of a solo lion (or very unlikely, of a lion pride). Nil Einne (talk) 05:44, 28 June 2017 (UTC)[reply]

Also as is nearly always the case, definitions are ~~fussy~~fuzzy at the boundaries. In particular, I don't think it's clear whether eating something hunted by your community as part of the community counts as scavenging or makes you a scavenger in that context. Is a lion cub a scavenger when it eats food caught by others in the pride in a hunt it didn't participate in? What about a male lion in a pride? (As our article mentions Lion#Hunting and diet group hunts in a pride generally involve the lionesses although male pride lions probably hunt more than we're previously appreciated.) When a lion (or lion pride) steals the prey of a hyena sure, it's often called a scavenger. For that matter if it steals the prey of a solo lion (or very unlikely, of a lion pride). You can look outside lions at e.g. animals that hunt for and bring home food for their mate. Once you start talking humans things get particularly ~~fussy~~fuzzy (edit: and not just in terms of predator/scavenger) but the human context being referred to tends to be much more like these ~~fussy~~fuzzy non humans animal examples than in a lions stealing from a hyena or another lion. (I wasn't totally happy with my earlier wording but it had been a while and since there were no replies decided it easier just to strike out the earlier response and repost it.) Nil Einne (talk) 11:02, 28 June 2017 (UTC)[reply]

I am surprised that predation wasn't mentioned in answers.

In the pig example, hunting is predation, it may be argued that animal husbandry is not

In the money example, the predator / scavenger / grazer categories just don't apply. Would you ask if a man that give money for a car is a carmaker, a mechanics, etc. ? The man is a consumer, ant that's it.

for you "many tiny" question, our predation article use the example of many ants killing and eating a much bigger prey, obviously each ant is considered a predator as well as the whole pack. See also meat ant, that takes for granted that they are indeed predators.

Bottom line: RTFM before asking questions

Gem fr (talk) 12:18, 28 June 2017 (UTC)[reply]

Can cross spiders (Araneus diadematus) change colour to match their surroundings?

Most cross spiders I've seen are brownish (or redish brown or greyish brown) in colour. However, I've seen a lot of charcoal-grey ones living on a bridge that had been painted that colour, and once saw a very bright yellow one on a hedge that contained leaves of that colour. Can the spiders change colour to match their surroundings? And if not, what accounts for the colour difference? Or are these difference species that I've mistaken for A. diadematusIapetus (talk) 13:35, 28 June 2017 (UTC)[reply]

each spider do not change colour to match its environment, but since Araneus diadematus can have a range of colours "extremely light yellow to very dark grey", the whole local population surely will follow Peppered moth evolution scheme, so that colour that give some adaptive advantage will prevail.

Gem fr (talk) 14:08, 28 June 2017 (UTC)[reply]

Yep, color of insects and spiders is hugely variable. To the extent that serious ID almost never uses color as a key characteristic. Here [10] is a nice scholarly overview of color in spiders, this [11] research is focused on color change during development of one (Araneae) species. SemanticMantis (talk) 14:45, 28 June 2017 (UTC)[reply]

Can hydrogen conduct electricity?

Science news had an article about Jupiter that said there is a theory that its magnetic field is caused by circulating electric currents in one of the planet's outer layers of molecular hydrogen. Does hydrogen conduct electricity? It does as a plasma but how about when it isn't ionized? Does it need a certain pressure? RJFJR (talk) 14:20, 28 June 2017 (UTC)[reply]

See metallic hydrogen. As is defining for a metal, this involves delocalized electrons (free to move) as a "cloud" within the metal, thus conduction. Yes, it takes a great deal of pressure. Andy Dingley (talk) 14:25, 28 June 2017 (UTC)[reply]

Is that the only time hydrogen would conduct? (No one has proved metallic hydrogen exists and it was proposed it might exist at the core of Jupiter but the layer described in the article is an outer layer so probably not sufficient pressure for metallic.) RJFJR (talk) 14:31, 28 June 2017 (UTC)[reply]

It is estimated that metallic hydrogen extend ~80% of the radius of Jupiter. That's fairly "outer" to me. Dragons flight (talk) 00:16, 29 June 2017 (UTC)[reply]

Here's a paper on the topic of metallization in Jupiter's atmosphere. Mikenorton (talk) 14:32, 28 June 2017 (UTC)[reply]

Thank you. RJFJR (talk) 14:42, 28 June 2017 (UTC)[reply]

If hydrogen is heated to a plasma (which some may say is not really hydrogen anymore because it is a different state of matter) that will conduct. As will any other element in a plasmic state. Aspro (talk) 21:57, 28 June 2017 (UTC)[reply]

But Jupiter's atmosphere isn't thought to be a plasma. Andy Dingley (talk) 22:20, 28 June 2017 (UTC)[reply]

Ionic hydrogen H⁺ conducts in solution. μηδείς (talk) 00:09, 29 June 2017 (UTC) (surprised that's a redlink. Maybe it should redirect to acid?)[reply]

June 29

Physics in different engine designs

2 engines have the same number of cylinders, engine displacement, stroke ratio (etc). One is shaped like a V with 2 cylinder heads while the other is shaped like an I with one cylinder head. Why is it that the I shaped engine has more power output in the same measured timeframe as the power output of the V shaped engine? 64.170.21.194 (talk) 02:22, 29 June 2017 (UTC)[reply]

@@ Line 340: / Line 340: @@
 = June 29 =
+== Physics in different engine designs ==
+engines have the same number of cylinders, engine displacement, stroke ratio (etc). One is shaped like a V with 2 cylinder heads while the other is shaped like an I with one cylinder head. Why is it that the I shaped engine has more power output in the same measured timeframe as the power output of the V shaped engine? [[Special:Contributions/64.170.21.194|64.170.21.194]] ([[User talk:64.170.21.194|talk]]) 02:22, 29 June 2017 (UTC)