Wikipedia:Reference desk/Archives/Science/2016 June 21

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June 21

When will a F-35 be fully ready? Why does the USS Gerald Ford take so long to deploy?

The article doesn't say on the first one and says she'll spend 2016-2019 doing.. something on the second. Why does it take 3 years between being commissioned into the Navy and being deployed? The 2005 article said the F-35 will enter service in... 2008. Sagittarian Milky Way (talk) 00:41, 21 June 2016 (UTC)

For the F-35, it depends on what you mean by "fully ready". The F-35B is currently at initial operating capability with the USMC and is deployed with VMFA-121. Deliveries are currently scheduled for the next 20 years; undoubtedly, the aircraft will receive upgrades during that time (it's the case with all other major aircraft). Service life is anticipated for another 35 years beyond that; undoubtedly, the aircraft will receive upgrades then, too. Full operational capability is currently forecast for 2021 per several search results, but that definition means (among other thing) "all units scheduled to receive a system have received it" -- is that a precursor to say that the aircraft is "fully ready" by your standards?

For the USS Gerald Ford, I'm doing more guesswork, but I suspect the long working-out process is because this is the USN's first new class of carrier in about 45 years (depending on where you want to snap the line on the USS Nimitz) — the gap between the Ford and the Nimitz is the same as that between the Nimitz and the USN's first fleet carrier, the USS Lexington. Per the Ford's article, there's a lot of testing to wring out all of the new tech so that future ships in the class don't have the same growing pains. — Lomn 02:03, 21 June 2016 (UTC)

When will testing finish on at least one variant and the first individual aircraft go "on duty" and have all its version 1.0 capability available to be used within about whatever the in service scrambling time's supposed to be? No taking an extra hour cause "Oh shit, it's not fueled, the non-trainers weren't supposed to be fueled for another 3 months, somebody get a fuel truck! Who here has finished training? Somebody arm the weapons! Find the guy with the other arming key!..." (however unlikely it is that we'd be threatened by something that needs an F-35 that early in the product lifecycle) Sagittarian Milky Way (talk) 05:55, 21 June 2016 (UTC)

The problem with the Ford deploying is... μηδείς (talk) 12:41 am, Yesterday (UTC−4)

Ah, I didn't realize the infobox's "status: testing, training" only means the A & C are testing and they don't have a B to spare from training (with the F-35 pilot cadre growing for decades and low immediate battlefield need and all). I guess it's happened already then. Sagittarian Milky Way (talk) 21:01, 21 June 2016 (UTC)

How to label a 0 distance in an engineer drawing

Let's say I have a starting raw material that looks like the first frame[1]. If I want the left half to be machined so that it's 10 units higher than the right half, I would draw it like the second frame.

But what about the case when that distance is 0? As in, I want the left half to be flush with the right half like the third frame? How do I draw that? Any similar example would be welcome. Johnson&Johnson&Son (talk) 01:44, 21 June 2016 (UTC)

have a look at these examples. I'm not sure there is precisely such a thing as you are asking, I think you would simply include the dimension on the side as normal and an axis line, to make it clear you could include a "right angle" mark in the corner of the object and then if you've drawn a straight line, it's assumed the line is straight. If you want to you could probably include a dimension on both sides, but I don't think it's necessary. You could draw a "ghost" of where the material you need removed is, with the original dimension. Vespine (talk) 02:22, 21 June 2016 (UTC)

Thanks for the help.

>I think you would simply include the dimension on the side as normal and an axis line

>If you want to you could probably include a dimension on both sides

I cannot dimension the left side against the bottom nor the right side against the bottom due to variation of the material. I would love to if I could, but I can't, hence the tricky problem.

I can only dimension the left side with respect to the right side. Johnson&Johnson&Son (talk) 03:11, 21 June 2016 (UTC)

Basically imagine the bottom as a rough irregular shape like this[2], impossible to reference against.

But I do have a nice flat repeatable reference surface, which is the right side. So I'm forced to rely on only referencing against it and nothing else. Johnson&Johnson&Son (talk) 03:16, 21 June 2016 (UTC)

It's rather unusual to mill both sides down independently when you want them both to be flush, as this process is likely to leave a discontinuity between them. But, if this is the only machining possible, an engineering note, perhaps with leaders pointing to both sides, might be in order. It could say something like "Mill both sides flush to each other and sand out any discontinuity." StuRat (talk) 04:40, 21 June 2016 (UTC)

Only the left side is milled. The right side is simply the shape of the stock. It's like a L channel.

How would the leaders look like? If you don't mind, could you please take a screenshot or even just make a rough paint drawing like mine would be fine.Johnson&Johnson&Son (talk) 05:58, 21 June 2016 (UTC)

That's a fairly useless suggestion if you are dealing with a professional machine shop, they could quite rightly charge you ten thousand dollars for that. The correct solution is to define the 'as is' surface as a reference plane, and then define the machined plane relative to that using the appropriate tolerance form and value, depending on what you need from the machined surface. http://www.gdandtbasics.com/parallelism/ would be a place to start, but you could just apply a flatness tolerance to the entire contiguous surface, ie bothte machined and unmachined parts. Greglocock (talk) 07:16, 21 June 2016 (UTC)

Here's some examples: [3] (straight leader lines, at low resolution, unfortunately), [4] (curved leader lines). The note can either be placed directly, or a number can be placed there, and circled. That number then references a note on either the side of that page or on another page. StuRat (talk) 13:59, 21 June 2016 (UTC)

Voltaic Cell

Hello, I am having trouble wrapping my head around the chemistry/physics behind lead-acid batteries. A textbook I'm reading tells me that in a battery (which is not hooked up to anything) maintains a potential difference (the cell potential difference, determined entirely by the chemical's redox potentials). It does so by having the following half-cell reactions occur on the electrodes, which again are not in contact (except by the electrolyte):

${\ce {{HSO4^{-}}+Pb->{PbSO4}+{H+}+2e^{-}}}$ (anode)
${\ce {{PbO2}+{HSO4^{-}}+3{H+}+2e^{-}->{PbSO4}+2H2O}}$ (cathode)

I have two questions:

• I would say, if the reaction were to theoretically start on the anode, then the anode reaction makes perfect sense; the reaction will produce two electrons that travel through the lead metal, and end up at the terminal. However, the positive and negative terminals are not in contact; my question is then, where do the 2 electrons come from on the cathode reaction? I have tried to come up with a reaction mechanism for how these two electrons could have perhaps come from the electrolyte, but I am stumped.
• Say the cell is at its cell potential difference, with the electrons at the anode terminal and the deficit of electrons on the cathode terminal. If the battery were to be attached to a load, would it be more correct to say that the battery "produces a force" that moves the electrons (from within the wire) through the circuit, or would it be more correct to say that the excess electrons originally on the anode terminal move through the circuit instead?

Thanks for the help. 74.15.5.167 (talk) 03:29, 21 June 2016 (UTC)

Does this answer your question ? The cathode and the anode ARE in contact, through the electrolyte. Vespine (talk) 04:35, 21 June 2016 (UTC)

Elecrons travel from anode to cathode through the load by wires (when there is one). The battery produces the electromotive force. Ruslik_Zero 17:23, 21 June 2016 (UTC)

Okay, but does that mean there is no chemical reaction that transfers these electrons through the electrolyte? That is, it sounds to me like, if the battery was not hooked up to any load or wires, then the anode could produce its chemical reaction but the cathode could not (as it requires the anode's electrons). Would that not imply that a battery on its own has no potential difference if it weren't hooked up to a load? 74.15.5.167 (talk) 23:59, 21 June 2016 (UTC)

The key thing to grok here is that electrons carry a LOT of charge. So an undetectable shift in the proportion of how many + ions and how many - ions are in a given part of a solution adds up to a huge change in voltage. This is how an electrolyte works - the ions can simply move back and forth. They can "brigade", too - if one negative ion moves a few angstroms one way, and the next one down the line does the same, and the one after that... all the way to the far end of the electrolyte, then the charge effectively moves long before any one ion could ever cross the distance. So in practice, you have an electrolyte with a certain voltage here and there within it, and that is putting an electromotive force on the ions, and you can measure that in aggregate without tracing a single completed motion that takes the electron from one end to the other.

Now the battery therefore can have a difference in charge without having any noticeable effect on how many electrons or lead ions are present at each end, because only an undetectably tiny fraction have to change before there is a voltage difference. Once there is a voltage difference, that difference is pushing back against the electrons, against the electrolyte, so that the equilibrium no longer favors any further reaction at either end. I think -- I might be saying more than I know about this part, because there's a distinction between discharging and recharging a non-rechargeable battery and simply having it sit on a shelf. I think this is because the equilibrium occurs at a very small scale, whereas discharging the battery substantially creates macroscopic (or at least microscopic) changes in its structure. Wnt (talk) 00:10, 22 June 2016 (UTC)

Freezing water and oil

What aspect has the solid obtained by freezing an emulsion of water and oil? Are the oil bubbles traped in ice? What is the difference from the case of two distinct layers of water and oil? Does the heat transfer between layers of water and oil play affect the structure of the solid obtained?--5.2.200.163 (talk) 15:36, 21 June 2016 (UTC)

It depends on the type of emulsion and how it is frozen. It seems pretty complicated. Here are two research articles about the freezing of oil/water emulsions [5] [6]. Here's [7] another that is freely accessible. SemanticMantis (talk) 17:05, 21 June 2016 (UTC)

Thanks for answer. I see that coalescence (chemistry) is mentioned in the summary of one of this articles.--5.2.200.163 (talk) 11:07, 24 June 2016 (UTC)

In the scientific name Sus scrofa L., what does the L. denote?

I would like to include the sub-species of wild boar found in the UK at the Wild boar article. This article states that it is Sus scrofa L. What does the L. denote? DrChrissy ^(talk) 20:20, 21 June 2016 (UTC)

Naming authority, i.e. described the species, and often who gave it the name that is being used. By convention, a simple 'L.' means the big guy, Linneaus himself. Kind of surprising at first how many things he got his mitts on, even from around the world. This is the simplest usage, there are all kinds of arcane variants supported by formal ICN/ICZN policy, but I don't see anyone other than systematists/taxonomists using those much. There are ways to describe who described it, who renamed it, and then how the genus was moved into another family, but that's way above my pay grade. Altogether a commonly confusing thing, one that's surprisingly hard to figure out if you can't just ask someone what's up :) SemanticMantis (talk) 20:23, 21 June 2016 (UTC)

Thanks and you deserve a prize for the speediest answer of the year! Looking back, I should perhaps have realised because the L. is not italicised; I had been thinking it was an abbreviation as part of the name, but obviously it would be italicised if it were. DrChrissy ^(talk) 20:30, 21 June 2016 (UTC)

Like my namesake, I strike quickly, especially on matters of meaning ;) SemanticMantis (talk) 20:37, 21 June 2016 (UTC)

References

I believe your aquatic namesakes can strike so quickly they make light (see Sonoluminescence)! How awesome is that! DrChrissy ^(talk) 20:48, 21 June 2016 (UTC)

And today I learned that there's a redirect from Linneaus. —Tamfang (talk) 08:32, 22 June 2016 (UTC)

Haha yeah I realized my typo eventually but then decided to rock the redirect from misspelling :) SemanticMantis (talk) 13:38, 22 June 2016 (UTC)