Wikipedia:Reference desk/Archives/Science/2013 March 27

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March 27

Are there any solvent of cyanoacrylate that does not dissolve polystyrene?

I had just found a bottle of dried cyanoacrylate. Is it likely to be still useful? I also wondered if I can use another solvent instead, because propanone dissolves polystyrene that I would like to glue.--Inspector (talk) 02:50, 27 March 2013 (UTC)

No, it isn't merely dried out, it has probably also polymerized. That is, cyanoacrylate glues don't work by merely drying out; there's a chemical reaction that occurs when they set, and the chemical reaction isn't really reversible. The idea behind cyanoacrylate (Super Glue, among other names) is that it spontaneously polymerizes when it is exposed to the humidity in the air. The monomers are dissolved in a highly volatile solvent (like acetone/propanone) so that they remain as monomer units; but as the acetone evaporates, the remaining monomer containing solution reaches the critical concentration and bingo: polymerization. This is a one-way trip: you don't get a solution of monomer units back when you dissolve it in acetone. Now, the polymer will soften some in acetone, often enough to break it up and remove it from, say, your skin. But it won't magically become monomer units again if redissolved in acetone. So, if you have a bottle of spoiled cyanoacrylate, the only way to get it back is to buy a new bottle. That one is toast. Regarding glueing polystyrene: you're also screwed there with any cyanoacrylate glue. There isn't, as far as I can think of, any solvent which will dissolve cyanoacrylate and not polystyrene. If you're trying to glue polystyrene to anything (either another piece of polystyrene, or another substance) you need a special polystyrene glue. Regular "Elmers" school glue should work too, but it usually doesn't hold very strong. Here is a good set of instructions on how to do it. --Jayron32 03:16, 27 March 2013 (UTC)

Thanks. I just got idea when I remembered some websites vaguely say about special types of CA that might be used on polystyrene.--Inspector (talk) 08:51, 27 March 2013 (UTC)

Flatulence

1) What is the average speed of a fart, in litres per second?

2) An unrelated question: why is a bad idea synonomous with the phrase "brain fart"? Plasmic Physics (talk) 06:05, 27 March 2013 (UTC)

2) A "brain fart" is an occasional undesirable result of thinking, just as a fart is an occasional undesirable result of digestion. StuRat (talk) 06:28, 27 March 2013 (UTC)

Our article flatulence is surprisingly good, referencing 5-375 ml for the typical volume - and, well, you know how long it goes on for. Wnt (talk) 14:32, 27 March 2013 (UTC)

Is it possible to control the pitch of a fart? What's the usual pitch range? Double sharp (talk) 15:51, 29 March 2013 (UTC)

It's hard to guess, since it contains many subharmonics, but the primary could cover he lower two thirds of the audible range. Plasmic Physics (talk) 03:59, 30 March 2013 (UTC)

Isn't the pitch proportional to sexual experience? — O'Dea (talk) 20:59, 31 March 2013 (UTC)

Heat released or absorbed in a reaction

I read these statements in an encyclopedia. Energy is conserved in chemical reactions. If stronger bonds form in the products than are broken in the reactants, heat is released to the surroundings, and the reaction is termed exothermic. If stronger bonds break than are formed, heat must be absorbed from the surroundings, and the reaction is endothermic. Earlier I was thinking just opposite. I think that the term 'exothermic' should be replaced by 'endothermic' and the term 'endothermic' should be replaced by 'exothermic' in the above italic statements. So, someone, please, explain the above statement. Concepts of Physics (talk) 06:05, 27 March 2013 (UTC)

I'm not sure why you think that's wrong, but our articles endothermic and exothermic explain it all. Rojomoke (talk) 06:13, 27 March 2013 (UTC)

Exothermic means means rejecting (pushing out) heat, in this case from hidden away in chemical bonds to becaome apparent in raising the temperature of the reactant/product mix - heat that will flow out into the surroundings. Endothermic means the temperature of the system is lowered in order to provide bond energy. Stronger bonds are so because they need less energy to make. Hence the statement you read is correct. It can be confusing to some because you often need to apply heat to break the bonds of the reactants in order to start the reaction off. But the heat you get back in an exothermic reaction is greater than the heat you put in. Wickwack 121.215.47.30 (talk) 06:23, 27 March 2013 (UTC)

The above explanation is very good, but try this one on for size as well: Bond strength is measured by how much energy you need to put into it to break it. Breaking bonds is always endothermic. Thus, by symmetry, forming bonds is always exothermic. If the bonds you break (the reactant side) are stronger than the bonds you make (product side), that means you need to put more energy into the reaction (because breaking is more energetic than making), so the whole process is endothermic. If the bonds you break (the reactant side) are weaker than the bonds you make (product side), then you get extra energy out, because the making step releases more energy than the breaking step needed to break its bonds, and the whole process is exothermic. Thus, if stronger bonds get made (products) the process is exothermic, while if the stronger bonds need to be broken (reactants) the process is endothermic. That's why the statement as written "If stronger bonds form in the products than are broken in the reactants, heat is released to the surroundings, and the reaction is termed exothermic. If stronger bonds break than are formed, heat must be absorbed from the surroundings, and the reaction is endothermic" is perfectly correct. --Jayron32 13:06, 27 March 2013 (UTC)

Others have discussed the science at play, but if you are hung up on the words, note that there is a bit of arbitrariness involved. See e.g. Exothermic#Contrast_between_thermodynamic_and_biological_terminology, which explains how the biological usage of these terms is basically the opposite of the thermodynamic usage. SemanticMantis (talk) 14:00, 27 March 2013 (UTC)

Car battery chargers with different amperages, what's the point?

I got a car/motorcycle battery charger for charging either 6V or 12V batteries with the choice between 2A, 4A, and 6A, and on the back there's a chart saying how many hours it will take (assuming the battery's 50% charged at the start). Obviously, when you select 2A or 4A on the device, it takes longer. Why would anyone select the lower options and voluntarily take longer? 20.137.2.50 (talk) 14:41, 27 March 2013 (UTC)

The chargers with smaller output cost less. Pay only for what you need. Battery chargers have different purposes.

(response to above unsigned line) This one charger has settings to choose between 6A (fastest) 4A (medium) and 2A (takes longest). 20.137.2.50 (talk) 14:56, 27 March 2013 (UTC)

If you take a car out of service for an extended period (e.g., a 6 or 12 month working holiday in another country), the battery may be flat and somewhat ruined when you get back. To prevent this, put the battery on continuous trickle charge. You need only a 2 Amp charger to do this, and most of the time it won't actually be delivering more than a fraction of that, unless the car security system draws a fair bit, which some do.

However, a 4 to 6 amp charger gives you some flexibility. If, say, you go to start your car and find the battery flat becasue you left the headlights on, connect a 4 to 6 Amp charger, wait 30 to 45 minutes, and your car will now start. Vehicles with larger batteries and larger starter motor draw (eg diesels) will need more amps or more charge time.

4 or 6 amp charging can be excessive for the small batteries used in light motorcycles that do not have electric start (50 to 125 CC class).

Wickwack 121.215.47.30 (talk) 14:52, 27 March 2013 (UTC)

OK, if the benefit of voluntarily choosing lower amperage and longer time is that it is less damaging to deeply drained batteries, then that is the answer to my question of why anyone would choose 2A. I wonder what situation makes 4A preferable to the still faster by a few hours according to my chart 6A. 20.137.2.50 (talk) 15:02, 27 March 2013 (UTC)

It is not a question of how deeply the battery is drained. You have misread or misunderstood if you think that. Ideally charging should be tapered off as a battery gets near full charge. Use the 2 Amp seeting if you are trickle charging a car battery - that is you are preventing the (fully charged) battery from going flat over time, as distinct from recharging a battery that has been flattended by some accident, such as leaving the lights on, or a defect in the electrical system. Use a lower charging rate on small motorbike batteries because any more will stress them - because they are samll, not because they are deeply discharged. Wickwack 121.215.47.30 (talk) 15:09, 27 March 2013 (UTC)

Thanks for the correction. The ad copy on the back of my thing says "fully automatic microprocessor-controlled battery charger 6 amp fast charge rate automatically adjusts charging rate" I thought that would be a one-size-fits-all that senses when not to give so much amperage, and was planning on setting it to 6 when I get home today, as the car was fine for the past 2 months since just replacing the battery and alternator until this morning when the low battery light came on and I took the other car to work but want to charge this battery up so I can get the 40 miles to the mechanic tomorrow as I suspect I got a short-lived alternator 2 months ago. 20.137.2.50 (talk) 15:24, 27 March 2013 (UTC)

Here are some references about battery charging current:

http://batteryuniversity.com/learn/article/charging_the_lead_acid_battery

http://www.evdl.org/pages/hartcharge.html

--Guy Macon (talk) 15:14, 27 March 2013 (UTC)

W boson in a hypothetical particle accelerator

I know that with a half-life of 10^-25 seconds, it is difficult to study a W in detail, but picturing one in some hypothetical super-strong particle accelerator seems theoretically valid. (Besides, who knows what devices people living shortly after the end of the electroweak era, composed of God only knows what kind of exotic matter, might have created?)

I have a feeling this house of cards will fall at an early step, but I'll ask all these at once for your convenience, and perhaps your amusement.

• Suppose a W+, whose temperature is just a little below the point of electroweak symmetry breaking, is accelerated in an evacuated chamber.

• Is it correct that the "temperature" of the isolated W+ is based simply on its velocity, per the Boltzmann relation?

• If it transitions to become massless at higher temperature, it should start moving at the speed of light, and be viewed as moving at the speed of light from any reference frame. Does that mean that there are two energy states, which I will ignorantly call W_+ and W*+, one of which is massed and one massless, and doing work on the particle involves the chance of promoting it to the more energetic state?

• Suppose we have a massless hot "W*+" moving toward a huge concentration of positive charge. It seems difficult to picture how relativistic electromagnetism applies to a positively charged particle moving at lightspeed... I get this notion that due to Lorentz contraction all the positive and negative charge in the cosmos (in the line of travel) is pancaked onto the W+, which makes it hard to figure, but there must be a mathematical treatment that avoids singularities?

• Anyway, let's suppose (dubiously) there's some way to do work against the "W*+" based on its charge. Presumably that should reduce its energy because it can't reduce its speed... since the charge doesn't go away, does that mean you could reduce its energy somewhere past the critical point, conceivably even reduce it to zero, before the "W*+" would decay back to a "W_+"? With a further push, could you abruptly stop it from moving forward at the speed of light and send it backward instead?

Wnt (talk) 15:03, 27 March 2013 (UTC)

I don't know much about the details, but I can make some general comments. First, Lorentz symmetry (the principle of relativity) implies that a rapidly moving particle has the same behavior as a particle at rest. For the same reason, a single fundamental particle can't have a temperature. You need a soup of particles with comparable kinetic energies and randomly directed momenta (relative to the center-of-mass frame) to have a temperature. Second, I don't think there's any phase of the standard model where all particles propagate freely at the speed c. At temperatures above the electroweak unification temperature the Higgs field no longer breaks the electroweak symmetry, but it doesn't disappear; there's a ton of energy in it (and every other field) at that temperature and all kinds of interaction going on, and no particle will get very far in any one direction at the speed c. If the zero state of the Higgs field is a pencil balanced on its point, and the low-temperature symmetry-breaking state is a pencil that fell on its side in a random direction, the high-temperature state is a pencil bouncing wildly off the walls. Third, electric charge doesn't make sense before electroweak symmetry breaking, so it's not a good choice for this example. Regarding charged massless particles in general, the gluons are a low-energy example in the standard model, and in gravity any massless particle (including a graviton) qualifies, but I have no good intuition for it either. Lorentz symmetry implies that you can't associate a time scale with a massless plane wave, and I think such waves don't self-interact. But if you have waves propagating in more than one direction there is a center-of-momentum frame and a nonzero total energy in that frame that gives you a time scale for interaction, even if they're individually massless. -- BenRG (talk) 17:31, 27 March 2013 (UTC)

Are you implying that if you take a W+ and heat it to the critical temperature, the positive charge is actually destroyed? Or does it become somehow "inactivated" but still conserved? Because the third option would seem to require it still be there on the W+, provided that particle hasn't decayed somehow to emit it somewhere else ... right?

I suppose one thing to check is... does the Higgs field as described have a definable velocity? Can a particle be moving "relative" to it? If not, what defines how it "bounces around" in relation to a single W suspended within it? Wnt (talk) 02:52, 28 March 2013 (UTC)

Electric charge is a certain combination of the U(1) fundamental gauge field and part of the SU(2) fundamental gauge field, which are still there at high energy, but the SU(2) is probably very different from its low-energy version (confining, like the strong force? This is the part I don't know much about). At any rate, that combination of fields still exists, but no longer behaves like electric charge, and there are a lot of other analogous combinations you could write down that happen not to be the electric charge at low energy, but seem to make as much sense at high energy. It's not an important point and I probably shouldn't have said anything.

In your second paragraph, are you talking about my wildly bouncing pencil? At high energy you're going to have a soup of particles/waves including Higgs bosons, and particles/waves have a state of motion. However the low-energy Higgs field is like the cosmological constant: it has a constant value everywhere which is locally Lorentz-invariant (so has no state of motion). -- BenRG (talk) 05:11, 28 March 2013 (UTC)

I suppose this is like an electron which modifies a field nearby it that normally is zero in every frame of reference... still, there are some things that seem hard to understand. The Higgs field now has a non-zero expectation value everywhere, but it is also described as being like a "Mexican hat". Why aren't some regions of space on the "north" side of the hat and others "south", "east", and "west"? And if the Higgs particle changes the field....... how? It has no spin, it is its own antiparticle, so how does the field know which way to change when it is nearby? Wnt (talk) 13:15, 28 March 2013 (UTC)

I happen to have a PhD in Grand Unified Theories, so a probably can answer some of those question better than Ben RG (Not a common situation). Let me start by answering BenRG's question. Both etectroweak and strong interactions are asymptotically free at high energy. That means the strength of the interactions decreases at higher energies/temperatures. That is the opposite of confinement, which requires an increasing strength of interaction (which happens at low energies/temperature). Something just came up here. I'll post another answer later. Dauto (talk) 17:56, 28 March 2013 (UTC)

The relation between atoms' kinetic energy and the energy levels of their electrons upon excitation

In other words, what really happens when an atom is excited, either by radiation or by collisions, or otherwise ? What are the mechanisms under which the transferred energy goes to exciting the electronic levels, the atom's kinetic energy as a whole, or both ? in the latter case, which is supposedly the most common case, what's the ratio between the two (elec energy levels & atomic kinetic energy) ? Any elaboration on these aspects will be great. BentzyCo (talk) 18:08, 27 March 2013 (UTC)

The kinetic energy (for a clound of atoms kinetic energy is translational kinetic energy and is its themperature) of an atom can be any value on a continuum from zero to whatever. However electron orbitals exist in discrete levels - they can only "jump" from one level to another.

When an atom encounters radiation, the radiation is best modelled as quanta - discrete packets of energy. If an atom recieves a quanta of energy matching the difference between one electron orbital and a higher one (not necessarily the next orbital in the series) then it absorbs the incoming quanta of energy by jumping to the higher orbital, without changing the atom kinetic energy. Conversely, if there is an orbital decay, a quanta/packet of radiation will be emitted. Decays occur at random intervals.

Atoms can also drop orbitals back towards the ground state by transferring the energy to their kinetic energy at random times. In this way kinetic energy increases in discrete steps. Once the electrons are in the ground state configuration, no more transfer to kinetic energy can take place. Conversely, an atom with sufficient kinetic energy (~thermodynamic temperature) to make a jump from one orbital to the next will do so at a random time, reducing its kinetic energy in a discrete step. If kinetic energy is insufficient (ie thermodynamic temperature is too low) to match an orbital change above the ground state, no transfer can take place.

Atoms can change/interchange their kinetic energy by collision with other atoms by whatever value satisfies the kinetic energy available already in the colliding atoms.

A atom at a given temperature ie kinetic energy below that required to change orbitals remains at that temperature indefinitely and does not require heat input. An atom at a temperature sufficient to jump orbitals will keep spitting out quanta of radiation at random intervals and thus looses energy. It requires never ending input of energy (either in variable amounts from collisions or in discrete jumps from incomming radiation of sufficent quanta size) in order to maintain temperature.

In a cloud of atoms, there is a random distribution of kinetic energy amongst the atoms. Some atoms will be more energetic than others. So in practice you don't see things suddenly change as you bring temperature up to a critical value - what you see is a gradual increase in radiation as you bring the temperature up. Similarly, with incomming radiation, you see a gradual apparently continously increasing temperature, as some atoms will be ready to jump an orbital level (and then to decay back to increased kinetic energy) and some won't.

I have not dicussed what happens with molecules or atoms combining at collision to form molecules - this is a further complex topic, where temperature is not solely determined by translational kinetic energy.

Wickwack 121.221.78.59 (talk) 00:30, 28 March 2013 (UTC)

Spacetime

Is it possible that time and space are separate but we cant physically do it? Or has that been unequivocally proven165.212.189.187 (talk) 18:36, 27 March 2013 (UTC)

It is possible that time is a purely human construct and not a physical dimension. On that basis, the answer to your question is yes it is possible. --TammyMoet (talk) 18:56, 27 March 2013 (UTC)

No, it is not possible, at least not on the basis of that article. Physics Essays is a journal that doesn't hesitate to publish crackpot ideas. I've managed to track down the original paper, and there's so many things wrong with it that I don't know where to start, but here's a small sample of the nonsense:

1. Most of it has no discernible meaning

2. The paper makes no predictions, whether testable or otherwise

3. The authors have absolutely no understanding of relativity: "the idea of time being the fourth dimension of space did not bring much progress in physics and is in contradiction with the formalism of special relativity". Time is not the fourth dimension of space; it is not a space dimension at all. Special relativistic formalism led to the idea that time is a dimension in the first place, because it transforms both space and time in a similar fashion.

4. The authors completely misunderstand Zeno's paradox, specifically Achilles and the tortoise. The basic paradox was solved by Aristotle 2400 years ago, and calculus put the solution onto a rigorous mathematical footing. Today, any 6th grader has the mathematical tools to plot distance vs. time for both Achilles and the tortoise, and see exactly why Achilles surpasses the tortoise. --140.180.254.209 (talk) 20:11, 27 March 2013 (UTC)

I don't understand the question. Time and space definitely are separate--you measure one with a clock, the other with a ruler. What you might be referring to is that according to Einstein's theories of relativity, time and distance are both relative. That means that if I'm on Earth and you're on a fast spaceship, we won't agree on how long the same event takes, or how big the same object is [insert bad joke here]. That doesn't negate the fact that both of us can take out a watch, take out a ruler, and measure both distance and length without any ambiguity.

You might also be referring to the fact that time is often considered a "dimension". First, even from a purely mathematical standpoint, time is different from the other dimensions of spacetime. The metric signature is either + − − − or - + + +, with the "special" sign always corresponding to the time dimension and the other 3 corresponding to the 3 space dimensions. Second, all the dimensions are "separate", in the sense that you can refer to them individually in any reference frame you want. --140.180.254.209 (talk) 20:11, 27 March 2013 (UTC)

I don't understand the question either. I wonder if the OP has read or misunderstood the article on spacetime. There's possibly some explanation of the nature of time at quantum level (Brian Cox's winding clocks), but I can't explain it because I don't really understand it. (Tammy might like to read The Unreality of Time for an older "theory", but it's philosophy, not science. ) Dbfirs 21:37, 27 March 2013 (UTC)

Matter and energy are in essence the same thing but you can't measure them with the same instruments. Electricity and magnetism were also thought to be different things until it was demonstrated by Maxwell that they were the same. Vespine (talk) 23:18, 27 March 2013 (UTC)

Electricity (more correctly electric field) and magnetism (magnetic field) are NOT the same. Maxwell showed how they interact to do various things. By suitable devices, energy can be transfered from one to the other. But that does not make them the same. It is often said that both make up the Lorentz force (the force on a point due to both) but this does not make them the same either. Neither is matter (more correctly mass) and energy the same. One can be converted into the other by certain means but that does not make them the same. I can convert food into poo, and other organisms can convert poo back into food, but please don't mix food and poo up.

Boffins talk quite correctly about mass-energy equivalence, but it is sloppy reading or sloppy use of the English language if you don't see the use of the word "equivalance" here as the mathematical sense of the word. I can say that there is an equivalence between heat energy and mass of a gasoline fuel, as from knowing one you can calculate the other with a standard formula, but that does not mean gasoline and heat are the same thing. This sort of sloppy language use seems to be very common in Ref Desk posts, but that doesn't make it right either.

Wickwack 121.221.78.59 (talk) 01:13, 28 March 2013 (UTC)

Electromagnetism:Originally electricity and magnetism were thought of as two separate forces. This view changed, however.... As for mass energy, well, again, you can't have mass without energy and you can't have energy without mass, wrt to the OP, he is asking could time and space be separate? One reply suggested that they could be separate since you can measure them with different instruments. Mass is measured with a different instrument to energy yet neither can exist without the other. I think the same can be said for electricity and magnetism. Apologies if I was sloppy with my language. But I still think my point in context is valid. Vespine (talk) 02:59, 28 March 2013 (UTC)

I have difficulty in determining context here as like others I don't understand what the OP was trying to ask.

You have directly quoted from the Wiki Electromagnetism article (History Section). But is the article right? You certainly CAN have an electric field without a magnetic field, and vice versa. Happens all the time, when, whichever one it is, is static (isn't moving). I have sitting in front of me right now a ring magnet that conveniently holds paperclips. There's no electric field around it. I've also got a 1.5 volt cell with nothing connected to it. There's an electric field around it. But absolutely no magnetic field. As to whether you can have mass without energy or vice versa - that's a bit harder to say. We'll have to wait until the boffins figure out matter at, or if required beyond, the sub-sub particle level. At any practical (measureable) level today, you CAN have one without the other, just as you can't realise the heat energy in gasoline unless you burn it in a heat engine. Wickwack 60.230.200.148 (talk) 03:29, 28 March 2013 (UTC)

By the time of Maxwell, people already knew that Electricity and magnetism were related phenomenons. That was obvious from Ampere and Oersted's work, and I don't think Maxwell added anything to that. What Maxwell really showed was that the potential-model of electromagnetism and the force-field-model of electromagnetism were equivalent. Someguy1221 (talk) 03:04, 28 March 2013 (UTC)

(ec) No, Tammy should not read The Unreality of Time unless he (she? it?) has a lot of time to waste. I've studied philosophy before, and most of it is intriguing, provocative, and mind-opening. The Unreality of Time, on the other hand, is pure bullshit dressed up as philosophy. The upshot is this: whenever anyone makes a claim about reality, the first question should be "does it fit with what we already know?" and the second should be "does it explain more than what we already know?" If the answer to either question is no, the theory is worthless. You can call it philosophy or religion all you want, but that doesn't save it from being wrong.

To add to what Wickwack said, you can unambiguously distinguish between matter and energy. You can also unambiguously distinguish between electric and magnetic fields. It makes perfect sense to say "the electric field here is 2 V/m, and the magnetic field is 0", or "an apple is matter, but a light beam is energy". --140.180.254.209 (talk) 03:18, 28 March 2013 (UTC)

Fair comment. I wasn't really recommending the article, and I accept your judgement that it's not even good philosophy. Dbfirs 07:49, 28 March 2013 (UTC)

UGH.. Mass–energy equivalence:In this concept, mass is a property of all energy; energy is a property of all mass. Saying gasolene has energy that is only released when you burn it is completely missing the point. Gasaolene has CHEMICAL energy which is released when it is burned, but ALL matter has/is energy as per E=mc^2. Electric field An electric field that changes with time, such as due to the motion of charged particles in the field, influences the local magnetic field. That is, the electric and magnetic fields are not completely separate phenomena; what one observer perceives as an electric field, another observer in a different frame of reference perceives as a mixture of electric and magnetic fields. I don't dispute the statements in these articles, if you do, feel free to find better references and change them. 05:18, 28 March 2013 (UTC)

All mass can be CONVERTED into energy per E=mc². The energy is not there anyway, just as the there is no heat coming out of a can of gasoline. In each case you have to DO SOMETHING to get the energy, and then you no longer have what you started with. Wikipedia is a valuable resource, but if I was to go through and fix all the wonky bits, I would need a team of hundreds if not thousands working full time to do it. Re your quote on Electric fields: The first sentence says the same as what I did, but the rest of it is a bit mucked up. An electric field cannot be percieved in another frame of reference as a magnetic field. If you are in a static electric field, it matters not whether you are still or moving, you still have only an electric field (unless you introduce a conductor). The nature of electric and magnetic fields as different things is clearly demonstrated by the fact that they decay off dramatically with distance, but electromagnetic waves propagate on a much more gradual inverse square law basis. Wickwack 60.230.200.148 (talk) 06:11, 28 March 2013 (UTC)

Look, we've gotten to the point where we're debating semantics instead of physics. Here's the physics. If you have some matter with mass m, you can turn it into mc^2 of any form of energy: photons, thermal energy, kinetic energy, whatever. If you have some energy, you can turn it into matter. If you have some energy, say potential, trapped inside an object, the object has greater inertia. That is, it's harder to move, and so more massive. General relativity treats rest mass energy and other forms of energy the same way; they all go into the stress-energy-momentum tensor. Does that satisfy your definition of "sameness"? If it does, that's fine; if not, also fine. (From 140.180.254.209)

You said: If you have some energy, say potential, trapped inside an object, the object has greater inertia...and is more massive. That's not right. A 1 kg object still has a mass of 1 kg whether it is moving or not, and whther it is raised to a height or not. I make no appology for delving into a discussion of language, because unless and until you understand the terminology, you cannot properly understand the subject, and anything either of us says will just get sillier and sillier - as shown by your statement I repeated in italics. You have confused potential and kinetic energy, and mixed up mass and kinetic energy. Wickwack 124.182.18.126 (talk) 07:29, 28 March 2013 (UTC)

Take two systems. System A is just like Earth, with box sitting on it. System B is just like Earth, but that same box is sitting at a slightly higher altitude. Everything else is the same. Well, System B is slightly more massive. That mass is not "in the box", but it has to be in the system somewhere. Someguy1221 (talk) 08:07, 28 March 2013 (UTC)

Nope. Both systems have the same mass - the sum of box and earth. Nothing is hidden. You can't use energy to increase mass with going through a conversion process. If the box System B is allowed to come to the Earth surface somehow (say because it was nudged over the edge of a platform, then the potential energy will be converted to kinetic on the way down. (Let's assume there is no atmosphere to much things up with drag) Upon impact, the kinetic energy will be converted into heat (and a bit of shock wave propagating in the eath until it too is converted to heat). So System B will see a temperature increase. But unchanged mass, unless the height, size of the earth, and consequent impact caused ionisation or atom smashing or some such, but that is just one of the ways to CONVERT energy into mass, and is in this example a special case. And a most unreliable one, because you'll more probably get a flash of EM radiation and a reduction in system mass. If you don't think so, explain your thinking with a bit more that just E = mc². Wickwack 121.215.45.143 (talk) 08:44, 28 March 2013 (UTC)

You seem to deeply underappreciate the mass-energy equivalence. I can use my boxes to turn a generator as they fall, and have that generator fire a laser that shoots photons into space. Those photons carry mass away from the Earth. Potential energy --> kinetic energy --> Light = energy == mass. It would be wrong to say the extra mass was in the box - it is in the Earth-box system. But it is there and you can't pretend it's not. You could also just look at Mass-energy_equivalence#Practical_examples to see where you're just flat-out wrong. Any change in the total energy of a system is associated with a change in the mass of the system. Someguy1221 (talk) 09:03, 28 March 2013 (UTC)

Well, I could say your good self deeply underappreciates the meaning of the word equivalence in this context. But calling each other dopey (albiet in a slightly more polite way) doesn't help much does it? Unfortunately the operation of a laser involves ionisation - as I said this is a method of CONVERTING mass to energy. In any case it is not the case that additional mass due to the box's potential energy was used to create the photon energy (and presumably you think that that could leave System B mass the same as System A, it is the case that some of the mass in the laser (part of the earth) got converted into photons. So System B then end ups with slightly LESS mass than System A (unless more energy from outside the system was added to the laser to allow the laser working gas to come back to ground state). The practical examples in the Wiki article do I think include one with a common misconception. Note that no references are cited for the examples given. Did the author make them up? I have come across the Grand Coolee Dam example before, in popular press, but not proper physics or engineering textbooks. As Wikipedia says, don't take Wikipedia articles as verified fact, use it to get new ideas and review the references. Have a carefull read of this: http://www.weburbia.com/physics/mass.html. Wickwack 121.215.45.143 (talk) 09:36, 28 March 2013 (UTC)

I honestly don't know if you're simply wrong, or arguing a semantic point. But let me try to clarify: are seriously telling me, that you believe, that if I show you two otherwise physically identical systems, and I heat one of them up by pumping energy in, that no mass was added to the system? That the increased thermal energy bears no associated increase in mass? That the box's inertia doesn't change? That the box's gravitational footprint doesn't change? I think you are misunderstanding your own link, Wickwack! It does not say that the mass does not increase. "In reality, the increase of energy with velocity originates not in the object but in the geometric properties of space-time itself. In the final analysis the issue is a debate over whether or not relativistic mass should be used is a matter of semantics and teaching methods." And many more quotes like it. Sure, the mass of the box does not change, but that is not the entire system. As I said, the energy of the system, the mass of the system, changes. You can go down this semantic road where energy does not have mass, but adding energy to a system alters its gravity and inertia. I call that mass, whatever word you want to use. Someguy1221 (talk) 09:59, 28 March 2013 (UTC)

In this case, semantics are important - it is at the root of the trouble. Yep - I'm saying that if you add mechanical forms of energy, or electrical energy, to a system, there is no inherent increase in mass. Neither does removing energy inherently reduce mass. An increase or decrease in mass does occur if there is a conversion. Conversion occurs in nuclear processes, emission and absorption of photons, ionisation, and the like. If the energy is applied in such as way as to add velocity to a mass, the time compression occurs to the mass as observed externally. The time compression produces an apparent (to the external observer) increase in inertia. Energy can be added without necessarily increasing velocity. The increased aparent inertia is seen in devices such as television picture tubes. The electrons are accellerated to something like 1/10th the speed of light over a distance of 300 - 400 mm. Tube design engineers take into account the apparent increase in inertia in calculating the beam deflection force required (delivered magnetically), slightly larger than would be expected from electron mass alone. There is a notional equivalent mass to any quantity of energy. Nowhere did I focus on the mass of the box alone - I referred to the system mass. Have a look at this one: http://henry.pha.jhu.edu/mass.pdf from John Hopkins Uni. Wickwack 58.170.139.239 (talk) 10:07, 28 March 2013 (UTC)

Someguy1221 is correct. There is no consistent way to maintain a separation of mass from energy gated by "conversion" operations. There is a real distinction between four-momentum and rest mass (the former is a vector, the latter is its length), but there is no sense in adding the rest masses of components of a system; it's the same as adding the lengths of a bunch of vectors that point in different directions. Someguy's System A and System B have different masses. The Wikipedia example of the hydroelectric power plant is accurate. -- BenRG (talk) 17:16, 28 March 2013 (UTC)

Merely repeating statements, as you've done, doesn't help much. The artcle has no references cited for the examples given. Do you have an authoritive reference for the power generation example? One that supports the tranport of mass over the grip, and not just one that calculates m = E.c^-2 where E is the output of the dam? Wickwack 124.178.52.113 (talk) 00:38, 29 March 2013 (UTC)

Wickwack, adding and subtracting energy from the total energy of an object always increases/decreases its mass. For instance this. -Modocc (talk) 01:13, 29 March 2013 (UTC)

Emission and absortion of light involves a conversion. Wickwack 124.178.52.113 (talk) 01:48, 29 March 2013 (UTC)

You wrote: "I'm saying that if you add mechanical forms of energy, or electrical energy, to a system, there is no inherent increase in mass". No. These bulk forms of energy transfer are primarily a consequence of the interactions of fundamental charges and photons. -Modocc (talk) 02:11, 29 March 2013 (UTC)

For incoming energy to atoms in the ground state (eg a noble gas) below the amount necessary to jump electron orbitals, what interaction occurs? Wickwack 120.145.140.148 (talk) 04:39, 29 March 2013 (UTC)

Electromagnetic induction which is mediated by photons. -Modocc (talk) 05:06, 29 March 2013 (UTC)

Which is the displacement of free electrons in metals and the like, which are not a noble gas, and need not be an absortion of energy anyway. Wicwack 58.167.232.210 (talk) 05:43, 29 March 2013 (UTC)

I suppose I could have been clearer when I said electromagnetic induction, since I mean the Maxwell–Faraday equation, because it describes the propagation of the light which gives rise to Van der Waals forces, covalent bonds, electrostatic attractions and even phonons. If you are asking what the possible interactions can occur within a medium or a reflector which does not absorb the radiation, my answer is the same, because its my understanding that this equation applies to this light too as it also interacts with the noble gas. Bound charges tend to be shielded though and have greater effective mass, nevertheless the atoms' charges do contribute to the radiation's propagation. Its very late here and I'm going to be busy tomorrow and this is off-topic to the OP's question, thus perhaps we can conclude and should hat/hab some of this at some point? -Modocc (talk) 07:14, 29 March 2013 (UTC)

Fair enough. I might re-introduce this as a question of my own in a few weeks - hope that is ok. It won't be good etiquette to pose a question and debate the answers though. Wickwack 124.178.141.34 (talk) 08:19, 29 March 2013 (UTC)

Look, Wickwack, I honestly don't know why you're disputing basic physics, but I'll show the evidence you're looking for. Look at Einstein's field equations, which I'll paste here for convenience:

$R_{\mu \nu }-{1 \over 2}g_{\mu \nu }\,R+g_{\mu \nu }\Lambda ={8\pi G \over c^{4}}T_{\mu \nu }$

The left hand side describes the curvature of spacetime. The thing on the right, T_uv, is the stress-energy-momentum tensor. It contains exactly what its name implies: energy density, momentum density, and stress. Do you see any distinction whatsoever between mass and energy? In fact, do you see mass anywhere in the tensor? You might be worried that Einstein's field equation describes the "very local" environment because it has derivatives, which is illogical if we're discussing the mass of a system. But consider this: in the low-energy limit, general relativity must reduce to the Newtonian limit, correct? In such a limit, there's still going to be no distinction between mass and energy, right? So in terms of gravity and inertia, general relativity does not discriminate.

If you're not convinced by this argument, see this paper. You claimed that a hotter object is not more massive, meaning harder to move. This paper claims the opposite: "many balk at the application to kinetic energy. Can it really be true that a hot brick weighs more than a cold brick?", and "We can thus tell our students with confidence that kinetic energy has weight,not just as a theoretical expectation, but as an experimental fact" --140.180.254.209 (talk) 07:44, 29 March 2013 (UTC)

I have a lot of trouble with the concept of a hot brick weighing (overlooking for moment that's not the right word - the paper actually has a distinction between inertial mass and gravitation mass. Weight by definition is determined by gravitation ) more than a cold brick, but am happy with a fast travelling brick being harder to deflect than a slowly travelling brick. That's like the electrons in cathode ray tube (oscilloscope tubes, TV tubes) example I mentioned elsewhere, which is well known to tube engineers. The electrons don't change their mass (which would affect what happens when they hit the screen) but to calculate the deflection force required for scanning, engineers need to, and do, take into account an apparent increase in electron inertia, expected as the flight time as seen by the deflection electromagnets is out compared to that "seen" by each electron, due to the relatavistic consequence of these electron bean going typically at 1/10th the speed of light or more. One of the few, and perhaps, only, situation outside nuclear reactors where relatavistic effects affect pratical Engineer's usual 3- or 4- place accuracy calculations. Wickwack 124.178.141.34 (talk) 08:36, 29 March 2013 (UTC)

I think I know what the Electric fields article was trying to get at. If you have a charge at rest, there's only an electric field. If you use a moving reference frame, there's an electric field and a magnetic field, because the charge is now moving. This is discussed extensively in classical electromagnetism and special relativity. Actually, electric and magnetic fields transform in a similar way as spacetime in special relativity, which shouldn't be surprising because Maxwell's equations were Einstein's inspiration for SR. --140.180.254.209 (talk) 07:17, 28 March 2013 (UTC)

You may or may not have noticed that I've been trying steer the discusion into talking about electric fields rather than point charges. Fields are more fundamental. You may get greater clarity if you think about electric fields (and magnetic fields). Consider a pair of parallel plates, large in area, spaced 1 m apart. One plate is held by some means a steady +1 MV with respect to the other. There is therefore an electric field strength at any arbitary position between the plates of 1 kV/mm. And I can measure that with a suitable instrument. There is clearly no magnetic field. If I move the instrument about within the field, say at 100mm/mSec in any x, y, or z direction, by some means that is non-conducting and no magnetic, I'll still always get 1 kV/mm. Now, tell me, if this instrument is also a magnetometer, what magnetic field will I get? Let's say I reapeat the experiment, this time moving the plates steadily apart while at the same time increasing the voltage on the plates at the same rate, maintaining the same 1 kV/mm electric field strength. What readings will I get off the dual electric field meter and magnetometer now? What if I hold the instrument at a fixed spot and move the two plates around, keeping them at the same voltage and distance apart? Think before writing please. Wickwack 124.182.18.126 (talk) 07:55, 28 March 2013 (UTC)

As 140.* said, Lorentz transformations mix electric and magnetic field components, and that's why people say you can't have an electric field without a magnetic field or vice versa in special relativity. Of course, in another sense of those words, you can.

If you move your electric field detecting instrument parallel to the electric field lines it will measure an unchanged electric field and no magnetic field, but if you move it in any other direction it will measure a slightly increased electric field and a small magnetic field. -- BenRG (talk) 17:16, 28 March 2013 (UTC)

(ec) Finally, some questions I can answer with mathematical precision! Suppose you have an instrument that measures both electric and magnetic fields, and you move it perpendicular to the electric field lines at 100 mm/ms, or 100 m/s. Then the new E' and B' you will measure are:

${\begin{aligned}&\mathbf {E} '=\gamma \mathbf {E} \\&\mathbf {B} '=\gamma \left(-{\frac {1}{c^{2}}}\mathbf {v} \times \mathbf {E} \right)\end{aligned}}$

where

$\gamma \ {\overset {\underset {\mathrm {def} }{}}{=}}\ {\frac {1}{\sqrt {1-v^{2}/{c}^{2}}}}$

Plugging in the numbers, I get E' = 1.0000000000000556 kV/mm and B' = 1.11e-9 Tesla, pointing perpendicular to the electric field. This calculation comes directly from the first equations in the article I linked to, classical electromagnetism and special relativity. If you hold the instrument still and move the plates, you get the exact same answers, because only relative speed matters. --140.180.254.209 (talk) 17:26, 28 March 2013 (UTC)

Yes, that follows from equations 2 and 3 in the Wiki article, simplified for zero initial magnetic field. I asked for three cases: a) fixed plates and moving probe, which you've answered, c) holding teh probe fixed, and moving the plates about, which you have reasonably inferred is really the same as a), and b) moving the plates steadily apart while changing the voltage on them so as to maintain the same field strength, which you did not answer. Can you please explain what measured values would occur in this case. Wickwack 124.178.52.113 (talk) 01:42, 29 March 2013 (UTC)

Since (by assumption) the field strength doesn't change as you separate the plates, the measured value is also unchanged. Did you have some reason for asking these questions? -- BenRG (talk) 04:03, 29 March 2013 (UTC)

Yes, I have two specific reasons. I would like 140.180.254.209 or someone else to come back on this before revealing both reasons - no offence to you Ben. However, nothing in the World is perfect, certainly not me, so I might therefore have made a mistake, though if so I don't know what it is (if I did, I would admit it and then shut up). So, if challenged, vigorously but without new logic (almost all this debate has been mere re-statement) I ask questions which hopefully might lead to a new insight, for me and you. I hope that makes sense. Wickwack 120.145.140.148 (talk) 04:34, 29 March 2013 (UTC)

Ben's answer is correct, with one caveat. If you move the plates apart, the plates carry charge, and moving charges generate magnetic fields due to the Biot-Savart law. Exactly in between the two plates, it turns out that the magnetic field is 0 due to the symmetry of the problem. (It might be zero elsewhere too; I'm too lazy to work out the exact answer, but it's just an area integral.) --140.180.254.209 (talk) 07:44, 29 March 2013 (UTC)

To clarify my question: I first thought about how to measure or explain time without using some aspect of space, and how to measure or explain space without using some aspect of time. If the state of t=0 is identical to t=1 then how can you be sure time passed? How do you differentiate 1km vs 1000km without using time to measure each?165.212.189.187 (talk) 16:01, 28 March 2013 (UTC)

There is an unavoidable circularity in the definition of time and distance and everything else, since physical concepts can only be defined in terms of other physical concepts. Physical theories can nevertheless make objective statements about the world by imposing more constraints than there are concepts. Then you can take n of those constraints as definitions of the n concepts, and the rest as testable experimental predictions. However, there's nothing forcing you to choose any particular set of n constraints as the definitions. For example, is F=dp/dt subject to experimental test, or is it merely the definition of force, or maybe of momentum, or even of time? Depending on the interpretation you pick, you may end up with different ideas of what a given experiment is really measuring, even though the experiment itself and the predicted outcome are both objectively unchanged.

That said, I think physical theories constrain the "reasonable" choices of definitions. Given the importance of Lorentz symmetry in physics, for example, it seems unreasonable to use a certain definition of time without using a symmetrical definition of distance. Professional physicists break this particular "rule" all the time, though, especially in relativistic quantum mechanics. -- BenRG (talk) 19:23, 28 March 2013 (UTC)

Detonators

Were radio detonators available in 1944? How much did they typically weigh? For comparison, how much did a typical clockwork fuze from the same time period typically weigh? Thanks in advance! 24.23.196.85 (talk) 22:47, 27 March 2013 (UTC)

Take a look at Pencil detonator for a simple timed detonator. These would have weighed only a couple of ounces and are literally the size of a pencil. Here's a patent for a radio controlled mine from 1942. Nikola Tesla had a radio control boat by 1898 and detonating an electrical detonator would be trivial compared to controlling a boat.Tobyc75 (talk) 01:18, 28 March 2013 (UTC)

See our Wikipedia article on the proximity fuze for good detail on what was developed during and before World War 2. Wickwack 60.230.200.148 (talk) 03:55, 28 March 2013 (UTC)

Well, I was asking specifically for the weight of a clockwork fuze -- not a time pencil, which is a type of chemical detonator. Also, it would be nice to know the approximate weight of a radio detonator -- I want to make sure that Alfred, my demolition man, doesn't get a loadout that's too heavy (he already has to carry a heavier loadout than anyone else in the team, because of the 21 lbs. of Composition C that he has to lug on his back). But thanks for the info about the radio-controlled mine -- that's just what I needed to know. 24.23.196.85 (talk) 04:51, 28 March 2013 (UTC)

Well, perhaps you should be more specific about what you want, so we don't have to guess. What do you mean by "radio detonator"? A proximity fuse? A radio controlled bob or mine? If the latter, a factory produced militarty specified device or an improvised device? People (terroists, resistance fighters, etc), have always been able to improvise things like model aircraft RC control devices to set explosives off remotely, and even cruder things requiring minimal technical skill, such as contacts glued to the speaker cone of an ordinary radio tuned to a quiet frequency. Wickwack 60.230.200.148 (talk) 05:47, 28 March 2013 (UTC)

By "radio detonator", I mean a device that sets off a demolition charge remotely by means of radio waves -- certainly not a proximity fuze. And I do mean a factory-produced device (such as those used by the OSS or the SOE), not an improvised device -- I don't think it's even possible to generalize about the weight of an improvised device because it varies so widely. 24.23.196.85 (talk) 06:13, 28 March 2013 (UTC)

The smallest WWII commercially available device that could work as a radio detonator would probably be the RBZ Receiver. See

http://www.cryptomuseum.com/spy/rbz/index.htm and

http://www.virhistory.com/navy/xmtr-ww2-port.htm (about 2/3 down the page).

The bottom part of Artillery fuze#Time fuzes has some info on WWII clockwork fuses. --Guy Macon (talk) 09:47, 28 March 2013 (UTC)

The RBZ was of course a militarised standard general coverage reciever for portable communications. Rather a waste to use it to explode bombs, which was definitely not its intended application. If standard commercial product was to be used as a radio fuse, it would make more sense, for cost, factory prioritisation, and operational reliability, for a modified model aircraft/boat radio control set to be used. Particularly in Britain, authorities made every effort to find a war production use for every factory, regardles of its peacetime role. For example, my mother as a young lady worked in a factory making toy electric trains, which involves making tiny electric motors. Mum wound the coils in the motors using some sort of hand operated winding machine. When the War started, toy train production stopped. The factory was converted to making an only slightly better quality tiny "motor" that was used with a tiny propellor as a generator to arm air-dropped bombs. Factories that made furniture, therefore having expertise in metal and plywood got to make parts for aircraft. None of this means of course that model RC derived equipment was used in WW2 as radio fuses, but it might be useful to research to find out or build into a plot. Ratbone 58.170.139.239 (talk) 12:01, 28 March 2013 (UTC)

First, find a radio receiver -- any radio receiver -- that [A] was available in WWII, [B] ran without a power cord, and [C] was smaller than the the RBZ Receiver. The various radios made for spies are a good place to look. If you can't find one smaller, then you have your minimum size and weight. --Guy Macon (talk) 12:39, 28 March 2013 (UTC)

A reasonable suggestion, I think. But choose one designed for morse code operation - they could easily set up to offer better immunity to false triggering from noise, and more easily adapted to close a contact to detonate the bomb. Small radios intended for spies in WW2 may not be a good idea. They were often regenerative or superegenerative. This means that the recievers put out a continuous radio signal of their own as a byproduct of their circuit technique. Regeneration and superregeneration gives very high sensitiviy with very simple compact circuits. The self radiated signal was judged to be acceptable where the opeartor was expected to only switch on at scheduled "listening out" times and switch off imediately the message was recieved. But the self radiated signal meant 1) the enemy could detect it and direction find onto it, and 2) the radiated signal could itself detonate some types of electric detonators. No good going bang just as you switched on! Also, these types were not very selective, meaning that false trigerring could occur from lightning, man-made noise, and transmitters on other channels. Last, but not least, they could require continual readjustment of the reaction control and tuning as the battery ran down, otherwise the sensitivity would drop off rather dramatically. Not a problem in recieving short messages, but may be a problem is left on for several hours. Best go for a radio described as "TRF" or better, "Superheterodyne". The RBZ is a sensitive selective superheterodyne state of the art for battery radios in WW2. Ratbone 124.182.145.28 (talk) 13:15, 28 March 2013 (UTC)

Would a "Sweetheart" (RCD Type 31/1) work? It says in the War Office catalogue that it's a TRF set, and the weight is given as 3 3/4 lbs (less than the RBZ set, but still heavier than the explosive charge itself). Oops, the 3 3/4 lbs. figure is for the total packed weight -- the weight of the actual receiver with power unit is 2 lbs. 2 oz., which is actually less than that of the explosive charge itself. 24.23.196.85 (talk) 00:59, 29 March 2013 (UTC)

Not a good choice. Despite the claim of it being TRF, the circuit given at http://www.cryptomuseum.com/spy/sweetheart/ shows it to be a regnerative 3-tube set of inferior quality, with a regenerative stage followed by 2 stages of low power audio amplification. It will have all the disadvantages of regenerative sets that I described above. It was designed to recieve broadcasts from the BBC etc in the then standard HF broadcast band. It is design to drive a piezo earphone, which makes conversion to closing fuse contacts not simple. A fully trained radio technician or engineer could certainly do it though. It appears that battery life and performance were considerably sacrificed to reduce battery size. In a radio fuse application that may require the receiver to be left on continuously for many hours, I would suggest using 3 telephone No. 6 batteries (which are physically much larger), in series for the filament supply, or several of the normal 4.5V batteries in parallel. Anyone using a raio set in enemy territory would have to buy batteies at frequent intervals. Radio tubes then drew a fair bit of current, while batteries back then were much inferior to what you buy now. It's difficult to tell from the manual, but it appears that the designers intended it to be run on the then standard bicycle headlamp battery. This would be readily available in enemy territory, and purchase would not attract attention like purchasing a specific radio battery would. Note that this radio is designed for use with an elevated 30 foot wire antenna. Performance would drop off rapidly if the wire is shortened for concealment purposes. However a simple modification would permit operation with a short wire, say a few metres long. NOTE: Since the method of tuning is somewhat uncertain, it would be necessary for the demolition team to set it all up at the site, with the actual detonator disabled, and have the transmitter send a test signal, with which the demolition team could tune the radio and verify that the fuse contacts close. Pre-tuning the reciever would not work. In contrast, pre-tuning an RBZ (a vastly better set technically in all respects) and locking the tuning in place with some sort of glue or wax before going anywhere near the site would be completely satisfactory. How far away does your plot require the transmitter to be? What are its concealment requirements? There is one advantage of the sweetheart set, and regenerative sets like it. Your team could take two. Use one as the transmitter - all that needs to be done make it transmitt is turn the regeneration control (the right hand knob accoding to the manual available at the crypto museum website) all the way up. Range would be rather poor and difficult to predict, though it should be reliable over a kilometre or so. Ratbone 121.215.43.129 (talk) 02:29, 29 March 2013 (UTC)

A more reasonable choice might be the so-called biscuit tin radio, MCR-1. See http://www.cryptomuseum.com/spy/mcr1/index.htm. This is a properly designed superheterodyne, far above the Sweetheart rubbish in ease of use and performance, but not up to the standard, cost, and bulk of an RBZ. It would have been well known to any SOE person. It also would require modification by a tech to fire a detonator, but the modification would be quite simple. Don't be misled by the size of the biscuit tin - your operator would not need all the various attachments eg for pwer main operation and would not need all the coil packs for different bands. It can work effectively with a metre or so of wire as an antenna. It can be trusted if pre-tuned and tested and the tuning locked with glue back at base before going anywhere near the demolition site. It should hold its' tuning and sensitivity as the battery runs down to a certain extent. The team can still use a Sweetheart as a transmitter to trigger it. As the Sweetheart's tuning is uncertain, the transmitter operator would just rock the tuning to and fro around the selected frequency until the bomb goes bang. Ratbone 120.145.25.156 (talk) 05:20, 29 March 2013 (UTC)

Thanks! The requirement here is to simultaneously blow up 3 parked Tiger tanks with 3-pound shaped charges, and the team doesn't have to move all that far -- just a hundred yards or so down the railroad tracks. Also, they won't be able to go back to the tank cazerne to retrieve the radio, even if it's left intact -- this is intended to be a feint to draw the Germans out, so once the tanks blow up, the whole place will be full of field-grays (exactly as intended). Would it be possible to use a single MCR to set off all 3 demo charges? 24.23.196.85 (talk) 05:54, 29 March 2013 (UTC)

I don't see why not. I, having an electronincs engineer background, know zilch about explosives and detonators, but the modification required for MCR-1 radios, which were designed to feed low impedance magnetic headphones, is to alter the output stage to operate a small relay. The contacts on this relay would be arranged to close a circuit, using the radio HT battery as a energisation source, to fire electric detonators and so set off the charges. I would think that all three detomators could be wired in parallel and fed from the same relay contact. As the required range is ony 100 m, things should be quite trustworthy, even if a Sweetheart is used as the triggering transmitter. Turn the Sweetheart reaction control all the way up, a quickish turn of the tuning knob past the selected frequency that the MCR is pre-tuned to, and up she goes! I presume the reason why you don't want to describe the team running a wire back 100 m is so that the Germans run around in indecision, and not follow the wire back, which would give an officer time to think maybe this is a feint - it being the natural tendency of a person in command to bark out some quick orders to keep the subordinates occupied and off his back while he has a think. Maybe the team should ensure the MCR radio is obliterted in the bang in order to increase the uncertainty and stress in those Germans. Ratbone 124.178.141.34 (talk) 07:50, 29 March 2013 (UTC)

Yeah, that's what I'm thinking -- rig up the MCR on the middle tank, and run wires (or maybe even explosive cord) to the other two. This will keep the Germans guessing, while the Maquis team is well on the way to its second diversionary objective (namely, blowing up a munitions train). 24.23.196.85 (talk) 04:52, 30 March 2013 (UTC)