Wikipedia:Reference desk/Archives/Science/2017 October 26

= October 26 =

Lead leaching out of metal
We sometimes fry things on a big slab of what seems to be carbon steel. Someone gave it to us from some plate steel place where they make stuff out of it, but not cookware. There are giant, thick sheets laying all over the place. It looks like steel or carbon steel or something. It rusts quickly but is nicely seasoned now and so is very non-stick. Do you think it's loaded with lead? Would lead leech out into the oil. Is there a way to test this? Anna Frodesiak (talk) 05:48, 26 October 2017 (UTC)
 * Your piece of steel plate is highly, highly unlikely to contain any lead. The only steel to contain lead is free-machining steel. This is a very specialised application for steel and there is no reason for free-machining steel to be rolled into flat plate. Alloying elements in steel, such as carbon and manganese, are tightly bound within the atomic lattice and they do not leach out of the steel. Dolphin  ( t ) 06:28, 26 October 2017 (UTC)


 * Thank you very much, Dolphin51. :) Anna Frodesiak (talk) 08:43, 26 October 2017 (UTC)


 * Also, with a properly seasoned iron surface, the food never touches the actual metal. --Guy Macon (talk) 15:57, 26 October 2017 (UTC)


 * Yes, but that doesn't mean the two don't chemically interact. Iron can dissolve in the oil, and some of that gets in the food.  This is actually considered a benefit of a cast iron pan, that they increase the iron content in your food. StuRat (talk) 16:33, 26 October 2017 (UTC)


 * Note carbon steel is the (current) most common material for making woks, Wok. May have been part of why the locals knew it was good for cooking. SemanticMantis (talk) 21:39, 26 October 2017 (UTC)

Thank you, all. Your helpful information led me to create and populate it and add the commonscat to the main article.

Now, I see and wonder if any or all of those should have the carbon steel parent category added. Thoughts?

Oh, and they want to get rid of the refdesk because it doesn't help other parts of the project. Phooey to that. You folks are wonderful and so helpful. Anna Frodesiak (talk) 01:31, 27 October 2017 (UTC)


 * The articles listed under “Crystal structures of steel” are all eligible to be included under “Carbon steel”. Incidentally, I notice the heading “Perlite (Steel)”. This spelling is definitely incorrect because perlite is something of interest in geology. In the context of carbon steel the spelling should be “pearlite” and is so named because of its resemblance to mother-of-pearl when viewed under a microscope. Can you tweak the spelling? Dolphin  ( t ) 07:10, 27 October 2017 (UTC)
 * Hi Dolphin51. Thank you! I've made the new cat, added the items to it, and speedy tagged the old one. I also fixed the article's commonscat.


 * Now, should I simply add carbon steel as a parent cat to Crystal structures of steel, or individually add carbon steel parent cat to Austenite, Bainite, Cementite, Ferrite (Steel), Ledeburite, Martensite, Pearlite‎? Best, Anna Frodesiak (talk) 00:33, 28 October 2017 (UTC)
 * I think it will be sufficient to add Carbon steel as a parent cat to Crystal structures. Dolphin  ( t ) 11:11, 28 October 2017 (UTC)
 * Done! Thank you, my friend! Anna Frodesiak (talk) 13:30, 28 October 2017 (UTC)
 * Small linguistic remark: You mean leaching.  "Leech" with the double-e is another thing entirely. --Trovatore (talk) 08:29, 27 October 2017 (UTC)
 * Ah, leaching, yes, of course. Thank you Trovatore. :) Anna Frodesiak (talk) 00:33, 28 October 2017 (UTC)

Mass to the Moon
I understand the moon is moving away from the earth by 4 centimetres every year, how much mass would have to be added or lost for it to stay in perfect orbit? JoshMuirWikipedia (talk) 12:59, 26 October 2017 (UTC)


 * Well, momentum is potentially unlimited, so if a well-placed alien has a sufficiently decent particle accelerator, and we wrap the Moon carefully in some sci-fi quality electrical tape to keep it from exploding, we ought to be able to knock it back where you want it with just a few protons. Of course, with current technology ... any method is impossible.


 * Also note this doesn't change the tidal acceleration; it would only be modifying the present orbit. The same should be true of any one-time mass impact. Wnt (talk) 13:31, 26 October 2017 (UTC)


 * Changing the mass wouldn't have any effect.
 * Sadly WP doesn't have a clear introductory article to this. Circular orbit is about the closest. Also Kepler's third law "The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit." was determined by observation, before Newton had worked out the gravitational theory behind it.
 * For a simplified circular orbit, the period of that orbit
 * $$T = 2\pi\sqrt{\frac{a^3}{GM}}$$
 * where:
 * $$T$$ is the orbital period
 * $$a$$ is the orbit's semi-major axis in metres (altitude from the centre of the Earth, in our simplified view)
 * $$G$$ is the gravitational constant,
 * $$M$$ is the mass of the Earth
 * So the Moon is moving away because it's slowing down. To bring it back, you'd have to speed it up. Andy Dingley (talk) 13:43, 26 October 2017 (UTC)
 * And the moon is slowing down because of gravitational drag due to tidal effects. Basically, some energy in the moon's motion is lost because it goes into distorting the earth's shape a bit.  That lost energy causes the moon to slow down ever so slightly which causes its orbit to drift outward ever so slightly.  This will not last forever, because other factors will lead to the moon-earth system becoming tidally locked so that there is no more drag on the moon.  -- Jayron 32 15:14, 26 October 2017 (UTC)
 * Wait, slowing down? No that's backwards.  The moon is speeding up.  (Higher orbits are faster orbits.)
 * It's speeding up because earth's spin (one rev/day. Fast) and the Moon's orbit (one orbit/month. Slow) are slowly converging. So the Earth is slowing its spin, while the Moon is speeding up its orbit.
 * ApLundell (talk) 15:56, 26 October 2017 (UTC)
 * Yes, of course. Damn sign conventions get me every time.  My bad.  -- Jayron 32 17:33, 26 October 2017 (UTC)
 * I don’t think “higher orbits are faster orbits” is right. In the formula given above by Andy, the time period T is proportional to $$2\pi a^{3/2}.$$ Speed S = distance per unit time, so T = const × distance/ S where distance travelled is the circumference $$2\pi a.$$ Equating the two expressions for T gives $$ka^{3/2}=a/S.$$ So higher radius a is associated with lower speed S. Loraof (talk) 17:39, 26 October 2017 (UTC)
 * No, he's right. It's simple orbital dynamics.  I've seen Buzz Aldrin (who did his PHD thesis on the matter) discussing it in layman's terms before; but the basic principle is faster = further out.  When you are in orbit, and you increase your forward velocity, you move out in orbit.  More kinetic energy = further from barycenter.  This is true in any rotational system (that's why electrons with more energy are at further distances from the nucleus of an atom, for much the same reason) Thus if the moon is moving outwards, it must be doing so because it is moving faster.  If you slow down, you move into lower and lower orbits; if you go too slow your orbit intersects the larger object and you crash into it.  -- Jayron 32 17:45, 26 October 2017 (UTC)


 * You’re describing speed between orbits. For objects in orbit with no externally imposed acceleration (other than gravity maintaining a given orbit), my math looks correct to me. Examples of orbital speed around the Sun: Earth 29.78 km/s; Mars 24.07 km/s; Jupiter 13.07 km/s. I was objecting to the statement “higher orbits are faster orbits”. Since the Moon is continually changing orbit, so to speak, part of its speed is not speed of orbit. Loraof (talk) 18:05, 26 October 2017 (UTC)
 * You know what, forget me again. I really shouldn't get involved in these problems.  I'm such an asshole.  -- Jayron 32 18:17, 26 October 2017 (UTC)
 * No, you’re not – you’re far and away the most helpful person on these ref desks. And your last comment was valuable in clarifying the difference. Thanks for all your work here! Loraof (talk) 18:24, 26 October 2017 (UTC)


 * If you add mass to Earth instead of the Moon and do it gradually then it might work. PrimeHunter (talk) 15:47, 26 October 2017 (UTC)
 * The moon is actually gaining energy from the rotation of the earth via the leverage of the tidal bulges (the earth's rotation drags these around so that they are ahead of the moon). Perversely (but in accordance with the virial theorem), for every one unit of energy transferred from the earth to the moon, the moon spends two units of energy in climbing higher: the one it got from the earth plus one from its own store of kinetic energy - so it ends up going slower. --catslash (talk) 16:47, 26 October 2017 (UTC)


 * As is always the case with questions like "what would happen to [gravitational phenomenon] if we added mass to [celestial body]", the answer is: it depends on how the mass is added. The above answer (in particular AD's) give the answer with the assumption that mass is instantaneously added with no change in any of the velocities at that instant. Tigraan Click here to contact me 18:43, 26 October 2017 (UTC)

Given
 * Moon's radius 1757.1 km
 * Moon's mass 7.342E22 kg

the Moon's center of gravity can be moved about 4 cm closer to Earth by adding a mass (7.342E22 x 0.04)/1732.1E3 = 1.70E15 kg on the side facing Earth, taking care to match velocities before contact. Continual deliveries of 2E11 kg every hour (the US annual waste production) might be a nice way to keep the Moon's orbit constant. Blooteuth (talk) 19:12, 26 October 2017 (UTC)


 * People think the moon is speeding up because the lunar month is getting shorter.  What is actually happening is that the moon is slowing down but our clocks are also slowing down because of the tidal drag (as measured by successive transits of the meridian by the sun, mean solar time).   The clocks are slowing faster than the moon, so it appears to be going faster. 92.8.218.38 (talk) 14:15, 27 October 2017 (UTC)


 * Kepler's third law requires that "the square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit." Hence we have $$\frac{T^2}{a^3} = \frac{4 \pi^2}{G(M+m)}\approx \frac{4 \pi^2}{GM} = \mathrm{constant}$$ where M is the mass of the Sun, m is the mass of the planet, and G is the gravitational constant. When this law is applied to any two masses, the constant of proportionality has, in turn, a value that changes with the system's total mass M + m.  Thus changing the Earth-moon system's total mass in any way has to change either the system's orbital period, the distance of their semi-major axis or both in order for Kepler's law to remain valid. With the problem at hand, if the angular momentum of the original Earth and moon masses are conserved then we also require that $$ \frac{a^2}{T} = k$$ where k is a constant. In other words, as their orbit about their common barycenter becomes more distant their orbital period is longer (as noted by others above).  Consequently, solving these two equations shows that increasing the system's total mass will arrest their separation and shorten the lunar months (such that Kepler's third law holds).  It's not a difficult calculation, but I haven't a lot of patience to punch in the numbers to determine how much mass would need to be added from afar each year, but you can have a go with it by first calculating the conservation constant for the current Earth-moon system as it now stands and plugging away into the following: $$ \Delta M = \frac{(2\pi * k)^2}{G \left( a + \Delta a\right)} - (M_1 + M_2) $$ -Modocc (talk) 04:59, 28 October 2017 (UTC)

Hydraulic motors
Why are hydraulic motors not more popular for vehicles? Hydraulic pump and motor pairs seems to be a very efficient way to transmit motive force from an engine to wheels. It eliminates friction losses from multiple gears, shafts and other moving parts in conventional transmission systems, and also weighs much less. I get the idea I'm missing information about one or more major disadvantages that explain why it's so rare. Roger (Dodger67) (talk) 19:01, 26 October 2017 (UTC)
 * Hydraulic linkages are indeed efficient and are employed in automatic transmissions. But your vehicle will need a prime mover i.e. a motor that converts energy from a source energy into mechanical energy. What source energy would you like to pay for? Blooteuth (talk) 19:25, 26 October 2017 (UTC)
 * I'd imagine an internal combustion engine driving a hydraulic pump would be the most obvious prime mover. The hydraulic linkage within a conventional transmission just one small component, I'm wondering why hydraulics are/were not used far more to basically eliminate almost the entire mechanical gear-and-shaft based drivetrain between engine and wheels? In recent years of course eletric drive has become far more efficient with hybrid IC/battery or even pure battery driven vehicles becomming more common. Hydraulics just seems to have never been considered a viable replacement for the conventional drivetrain. Roger (Dodger67) (talk) 19:43, 26 October 2017 (UTC)


 * Don't confuse hydrostatic transmission with hydrodynamic. Hydrostatic transmissions (pump and motor) are limited in power, speed and efficiency. They're mostly used when controllability is important, or high force vs. speed. They can also make accurate positioning mechanisms, as well as continuous rotation.
 * Hydrodynamic transmissions are those involving fluid couplings, torque converters and the transmissions of diesel-hydraulic locomotives are quite different. They can transmit high powers (multi-thousand horsepower) at high speeds, and are lighter, simpler and more compact than comparable diesel-electric transmissions. Andy Dingley (talk) 20:11, 26 October 2017 (UTC)


 * One problem with hydraulics is that they behave differently at different temperatures. Fluid viscosity changes, volume changes, etc.  Thus, a cold vehicle would drive very differently than a hot one.  There are ways to compensate for this, within limits. StuRat (talk) 03:08, 27 October 2017 (UTC)


 * Hydraulic fluids in power transmission systems operate at a fairly constant temperature, set by their cooling systems and thermostats. They are heated by use plenty to rise above ambient. If they're in a cold climate, they may be pre-circulated beforehand, just to warm them. Andy Dingley (talk) 13:58, 27 October 2017 (UTC)


 * Yes, and preheating is fine for, say, a crane, but car owners neither want to wait to preheat the hydraulics nor pay to keep them heated at all times. So, one point against hydraulic cars. StuRat (talk) 22:45, 27 October 2017 (UTC)


 * There are no hydrostatic cars. Nor are there going to be. Andy Dingley (talk) 22:56, 27 October 2017 (UTC)


 * Agreed. Of course, if we are looking at this from a historic perspective, then hydraulics might have worked better with steam engine cars, since they also require preheating. StuRat (talk) 23:03, 27 October 2017 (UTC)
 * What possible benefit would a hydrostatic transmission convey to a steam-powered vehicle? They already have the advantages of a hydrostatic drive (the ability to generate considerable torque from zero speed, without needing any transmission) as they are. If you can cite all these "hydrostatic transmission steam cars", then please do so. Andy Dingley (talk) 10:01, 28 October 2017 (UTC)

I found Human Friendly Transmission, used by Honda in a few motorcycles. Roger (Dodger67) (talk) 06:16, 27 October 2017 (UTC)


 * That is a continuously variable transmission (i.e. a continuous-ratio gearbox), which makes up only a small part of the drivetrain. An important part of OP's question is why isn't most / all of the drivetrain made by hydro links, which presumably have less friction (hence power losses) than mechanical drivetrains. (I do not have a clue.) Tigraan Click here to contact me 11:26, 27 October 2017 (UTC)


 * In case there's some confusion, Roger is the OP. Nil Einne (talk) 11:37, 27 October 2017 (UTC)