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February 5

Water decapitiation or being squashed like a pancake?

Suppose the US has the capabilities to construct a 5,000-ton (but only about 1000 tons' worth of space available to the crew) submarine that has a super-strong titanium hull capable of withstanding great pressure. Suppose that the sub is at the bottom of Mariana Trench, and water is leaking through a 5 mm hole in the ceiling. There is only enough food and water supplies to last the single submariner another 24 hours. He has two options -- to wait until the sub fills up and die due to drowning in the cold water and the immense pressure, or instead position his head underneath the leak so the rapid leak could kill him. Now he doesn't know how quick the leak is and how fast the water is travelling, so he hesitates. Can somebody help him out by calculating the water's actual velocity, and the water's required velocity at which the submariner's head will get punctured instantly and die? --Sp33dyphil ©^hat_ontributions 00:05, 5 February 2012 (UTC)[reply]

Note that the rapidly moving water will also rapidly enlarge the hole, so they wouldn't need to get bored waiting to die. :-) StuRat (talk) 00:10, 5 February 2012 (UTC)[reply]

Suppose all the dimensions stay the same. --Sp33dyphil ©^hat_ontributions 00:16, 5 February 2012 (UTC)[reply]

This sounds like a homework problem, no? Well, it would be a good one, and I don't use numbers. We could of course calculate the water pressure at that depth, but the Mariana Trench article provides it:

109x10 6

pascals (N/m^2). Pressure is force per area, and the area into which the force is being applied is a 5mm hole and

A= πr 2

, which gives the force of the water flowing in. To break a neck, I'll just use the Hanging#Long_drop article for the heck of it: 5600 N. I get 2000 N from the water stream (full calculation left as exercise for the reader). Note that the smaller the hole, the less force, so you might want to widen it to a couple cm before you go a-suicidin'. SamuelRiv (talk) 03:02, 5 February 2012 (UTC)[reply]

What would the water velocity for both cases be? --Sp33dyphil ©^hat_ontributions 03:13, 5 February 2012 (UTC)[reply]

The velocity will be affected by three things: 1. The orrifice effect as the water transitions from the sea into the hole, 2. The viscous drag while in the hole, and 3. the orrifice effect as the water transitions from the hole to the air inside the hull. You can get 1 & 3 from tables in hydraulics and/or fluid dynamics texts, and (2) can be calculated by applying the Fanning-Darcy equation arranged to solve for flow given the pressure drop, length of the hole (= hull thickness), and the viscosity of water at the temperature that exists at that depth. Fanning Darcy has 2 forms - one for laminar flow, one for turbulent flow. You need to first calculate the Reynolds Number, dependent of hoile dimensions and viscosity, to decide which Fanning Darcy to use.

So, you need to know the thickness of the hull. Unless you think titanium is so strong the hull is paper-thin - not realistic. Keit124.182.20.59 (talk) 03:43, 5 February 2012 (UTC)[reply]

If the hull is 2 cm thick, what's the velocity? --Sp33dyphil ©^hat_ontributions 03:45, 5 February 2012 (UTC)[reply]

Look up "Reynolds Number", look up "Fanning-Darcy equation", and look up a good text for tables on the two orrifice efects, and do the math. I'm happy to point you in the right direction, but not to do your homework. Keit124.182.20.59 (talk) 03:49, 5 February 2012 (UTC)[reply]

Naive velocity estimate is 400 m/s. SamuelRiv (talk) 04:01, 5 February 2012 (UTC)[reply]

This is not h/w. I was just curious. And wow, 400 m/s! --Sp33dyphil ©^hat_ontributions 05:19, 5 February 2012 (UTC)[reply]

And (assuming my maths is right, which it may very well not be), it will only take 159 seconds or thereabouts to fill your 1000 M³ space with water anyway, with it coming in through a 5mm diameter hole at 400 m/s. I don't think that food and water supplies are going to matter much here. AndyTheGrump (talk) 05:55, 5 February 2012 (UTC)[reply]

I suspect the 400 m/s flow is just a guess. But assuming it's right, 400 m/s thru a 5 mm hole is 0.00785 m3/sec. That will need 35.4 hours to fill the 1000 m3 sub. Keit58.170.173.63 (talk) 07:08, 5 February 2012 (UTC)[reply]

Er, yes. I've probably got this wrong, on consideration. Though I'm not convinced you're right either. Maths was never my strong point... AndyTheGrump (talk) 07:20, 5 February 2012 (UTC)[reply]

~35 hours is correct, given the assumptions. Your value, Andy, would require the hole to have 800 times the area it does. As for the flow rate itself, if we are allowed to ignore friction, I believe 464 m/s is a more accurate value. Although perhaps Sam's naivete operates on a higher level than my own, and he is able to do guess the affect of friction. Someguy1221 (talk) 07:38, 5 February 2012 (UTC)[reply]

Er, yes. Third time lucky, and I get the same answer: 35.37 hours, give or take. Never ask a bloke with a social science degree to design a submarine ;-) AndyTheGrump (talk) 08:28, 5 February 2012 (UTC)[reply]

400m/s was not a guess, but calculated using the hole bore volume to get the effective mass being ejected, without consideration for hydrodynamic effects. The water that actually exits will be very much more of a spray than a jet because of its turbulent speed and diffraction off the hole edges, and the work of displacing air will slow down the flow soon, but this is a nice upper limit. SamuelRiv (talk) 22:37, 5 February 2012 (UTC)[reply]

"mass being ejected without consideration of hydrodynamics"??? I would have thought that hydrodynamics, ie.e, Fanning-Darcy equation, is at the core of the calculation - please explain. Displacement of air can be neglected: Starting air pressure inside = atmospheric sea level pressure = 1 Bar, and the water pressure at the bottom of the Mariana Trench = 1000 Bar. Therefore, it's not until the air in the hull is compressed to less than 0.1 % of its volume that its presure will be comparable to the sea. Don't forget to allow for the Ratio of Heats for air (1.404). Keit60.231.241.252 (talk) 01:18, 6 February 2012 (UTC)[reply]

Hm, for the most basic approach I would have used the momentum change induced on the water by the pressure force -- i.e. p*A=v*dm/dt=v*density*A*ds/dt=density*A*v^2 <=> v=sqrt(p/density). For density=1e3kg/m^3 and p=1.1e8 N/m^2 that would be v=331.7 m/s. And from that you'd get the mass flow rate via v=ds/dt=dV/A/dt=(dm/density)/A/dt <=> mass flow rate=dm/dt=A*sqrt(p*density)=6.5 kg/sec. No? Olaf Klischat (talk) 18:43, 9 July 2013 (UTC)[reply]

I don't think displacement of the air can be neglected, since 400 m/s is faster than the speed of sound in air. In other words, it's a supersonic jet ;) Gandalf61 (talk) 08:41, 6 February 2012 (UTC)[reply]

Good point, I didn't think of that. However, upon thinking about it, the ratio of sea pressure to starting air pressure is so great (1000:1) it still won't matter. What might matter is that the resulting shock wave in such a confined space might knock out all the humans forthwith. Though I don't accept (without evidence) that the 400 m/s value has been correctly estimated anyway. Keit121.221.76.67 (talk) 12:42, 6 February 2012 (UTC)[reply]

Bernoulli equation has the water velocity from V=0m/s and p=1000bar to p=1bar and V=450m/s. --145.94.77.43 (talk) 03:27, 8 February 2012 (UTC)[reply]

Why should this submariner worry? Nitrogen narcosis is a nice euphoric way to go (as the pressure increases). If he is on-board with a mixed crew - all the better. --Aspro (talk) 23:17, 8 February 2012 (UTC)[reply]

What kind of spectrum is this?

Hi. One day while glancing over the surface of the water in a partly-full water bottle in a room with windows at the back and fluorescent lights over top, I noticed an unusual spectrum. At least part of the spectrum was likely from the fluorescent lights, but probably not the part you'd expect, since I observed the fluorescent spectrum in water to be very similar to the natural white light spectrum. I will list the colours of this spectrum in order, but keep in mind that italic text signifies what I call "narrowband", bold what I wrote down as the thickest part of the spectrum, taking up almost half of the entire spectrum, and underline for colours seen very narrowly at the edge of some transition, possibly into fluorescent lighting or ambient ceiling.

Tan
Pink
Rouge
Magenta
Violet
Indigo
Navy
Blue
Green
Teal
Lime
Tan
Haze
Orange
White
Orange
Red
Mahogany
Violet
Magenta
Venusian
Clear

Now, any idea what this spectrum may consist of? There did not be any objects directly contributing, unless there was a ubiquitous green backpack or other container within my line of sight. Thanks. ~AH1 ^(discuss!) 01:04, 5 February 2012 (UTC)[reply]

How in the world did you come up with those color names? Looie496 (talk) 02:52, 5 February 2012 (UTC)[reply]

You should ask Roy G. Biv. Edison (talk) 03:27, 5 February 2012 (UTC)[reply]

It's hard to picture an image from such descriptions, but the first non-bold colors seem to match those of the supernumerary rainbow depicted in the article. Reportedly this has something to do with interference... but I can't really speculate exactly how that appeared in a water bottle. Wnt (talk) 03:58, 5 February 2012 (UTC)[reply]

The normal spectrum sequence is seen with oil films on water surfaces due to interference within the oil. Your observed effect could be due to viewing this through another oil film on the inside surface of the bottle. This could thru addition provide more colours. Keit124.182.20.59 (talk) 03:59, 5 February 2012 (UTC)[reply]

In other words, AstroHurricane is not looking at a "spectrum," but rather a "colorful array of refracted and diffracted light." The root cause for such a colorful display is the same as prismatic diffraction - optical dispersion - but whatever was causing the diffraction was not high-grade, precisely-ground polished optical glass. Consequently, you see a muddied mix of a lot of different light, including some regions where clean spectral lines might have been visible. AstroHurricane, perhaps you can build your own spectrometer, using these ordinary house-hold materials, and let us know if your spectrum looks any cleaner. If you have access to a laboratory-grade prism, you can build an even better spectrometer, and should get even cleaner results. Nimur (talk) 16:37, 5 February 2012 (UTC)[reply]

Earth spin and distance to the sun

Sir, suppose if the spinning speed of earth (around its own axis)is doubled what will happen to distance between the sun and earth. will it increse or decrease? — Preceding unsigned comment added by Tobyaickara (talk • contribs) 12:55, 5 February 2012 (UTC)[reply]

Why do you think it would change? I can see no reason why it should. AndyTheGrump (talk) 15:21, 5 February 2012 (UTC)[reply]

Perhaps you should read tidal locking and orbital resonance. The orbit does affect the axial rotation in a very complicated way; the two parameters are, effectively, "coupled oscillators." Objects with larger orbital axis are much more weakly tidally-locked, effectively meaning that the time-constant for the resonant locking imcreases very significantly. We could go through calculations that are outlined in the articles I linked, if you have questions; but the OP will need to specify some details if they want a mathematical answer. As I always mention when people ask questions about cosmological hypotheticals: in order for us to answer correctly, you will need to be more specific: how would the Earth's orbit or axial rotation change speed? For example, would it be due to an impact with another object? In the absence of a cause, conservation of momentum - specifically angular momentum - tells us that there should be no change without outside interference.

Anyway, here is the equation that defines the most important part of the original question: the relation defining the time constant of axial rotation resonantly locking to the period of orbital revolution. Note the dependence on the semimajor axis of the orbit. If you want a more rigorous, less approximate form, I recommend de Pater and Lissauer's Planetary Science text, which covers orbital parameters very extensively and mathematically. Nimur (talk) 16:27, 5 February 2012 (UTC)[reply]

Another effect might be that the Earth would lose some of it's atmosphere, which would reduce the Earth's mass. However, I wouldn't expect this to be significant (but either would the tidal locking, due to the Sun's distance from Earth). StuRat (talk) 23:12, 5 February 2012 (UTC)[reply]

These effects are small (especially since there is no rotation–orbit resonance for the Earth), so, initially, and to a good first approximation thereafter, AndyTheGrump's intuitive reply of "no effect" on the year or distance if the day length was to be magically halved seems reasonable. In reality, as Nimur explains, magic doesn't happen. Dbfirs 22:47, 6 February 2012 (UTC)[reply]

Is amphibia a paraphyletic group?

Kinda related to the invertebrate question, seems like invertebrates are a paraphyletic group since it does not include all of the descendents of invertebrates (like vertebrates), so is this the same for amphibia? I know we evolved from tetrapods, which we are part of, but did we at one point evolve from a lifeform that was for all intents and purposes, an amphibian? ScienceApe (talk) 17:44, 5 February 2012 (UTC)[reply]

See Pederpes, which we describe as "Amphibia sensu lato". In Linnean taxonomy it is an amphibian, but in cladistic taxonomy it is a tetrapod. Other fauna from Romer's Gap may also be interesting to you. Wnt (talk) 17:53, 5 February 2012 (UTC)[reply]

"Invertebrata" is polyphyletic, not paraphyletic.^{See note} And define "amphibian". Do you mean living amphibians (Lissamphibia) or Amphibia sensu lato (all non-amniote tetrapods - Temnospondyli, Lepospondyli, and Lissamphibia, etc.)?

Amphibia sensu lato (i.e. including all extinct forms) is paraphyletic, since it does not include Amniota (reptiles, birds, mammals, including extinct forms).

If you meant the Lissamphibia alone. Quick answer is we don't know. See Labyrinthodontia#Origin of modern amphibians. Theories range from a monophyletic Lissamphibia derived from Lepospondyli; to separate origins of Caudata from Temnospondyli; to separate origins of caecilians from other amphibians sensu lato closer to amniote ancestors or even porolepiform fish (!).

Also see Tetrapod#Phylogeny. Note location of Amniota and Lissamphibia.-- OBSIDIAN†SOUL 18:48, 5 February 2012 (UTC)[reply]

Note: the traditional classification of Invertebrata included [macroscopic] animals from more than a dozen different phyla while excluding common ancestors ("Insecta" and "Vermes"). It even included plants! This is not even taking into consideration the problem of whether Metazoa (Animalia) itself is monophyletic or polyphyletic. But yeah I guess, Invertebrata can also be said to be paraphyletic in relation to vertebrates, but that's a bit superfluous. Like assembling a doll with three arms, two heads, no body, and five legs; and then detaching one leg because it was "different". :P -- OBSIDIAN†SOUL 19:57, 5 February 2012 (UTC)[reply]

See invertebrate. The first sentence of the second paragraph says, "Invertebrates form a paraphyletic group." ScienceApe (talk) 23:14, 5 February 2012 (UTC)[reply]

See above note (I have removed the <small> tags). In Linnean taxonomy, non-vertebrate animal phyla included only two groups: "Vermes" and "Insecta", and they included a mishmash of organisms (again even including Volvox, which is under Plantae). When Lamarck introduced the formal group "Invertebrata", he divided them into "Mollusques", "Crustacés", "Arachnides", "Insectes", "Vers", "Radiaires", and "Polypes". All of which do not include a common ancestor. Thus these groupings are polyphyletic. These were the only instances where Invertebrata was treated as a formal group. In the modern sense, the term "invertebrate" is completely informal with a definition that has become simply "all metazoans, excluding Vertebrata". It is thus paraphyletic to vertebrates, provided that you accept a monophyletic Metazoa.

And that article still needs a lot of work, which illustrates exactly how problematic the grouping is.-- OBSIDIAN†SOUL 02:21, 6 February 2012 (UTC)[reply]

It's important to distinguish between amphibia and amphibians. In modern biology amphibia names a monophyletic group which includes amphibians as well as reptiles, birds, and mammals. Similarly reptilia names a monophyletic group which includes reptiles, crocodilians, and birds.— Preceding unsigned comment added by Looie496 (talk • contribs) 00:47, 6 February 2012‎ (UTC)

Equation explanation

Can someone explain for me how to use this equation? 4π²r³/T² = G(m₁ + m₂). Thanks!Pendragon5 (talk) 19:48, 5 February 2012 (UTC)[reply]

It's a form of Newton's law of universal gravitation as it applies to a body in orbit. G is the Gravitational constant and m1 and m2 are the masses of two bodies, r is the distance between the centers of the bodies, π is 3.14159... and T is the orbital period (i.e. time for the two bodies to return to the same relative position). An alternative form of this same equation is at Orbital_period#Small_body_orbiting_a_central_body. --Jayron32 20:44, 5 February 2012 (UTC)[reply]

What is / sign, the one right in front of T stand for? And yes finally, this is the one i'm looking for. I just didn't understand it at first.Pendragon5 (talk) 20:57, 5 February 2012 (UTC)[reply]

Division. 4 times pi squared times r cubed divided by T squared. — Lomn 21:03, 5 February 2012 (UTC)[reply]

WOW, there should be a bracket () around 4 times pi squared times r cubed so people can understand it. And by the way, what number is Gravitational constant? I look in the article i still don't get it. Can someone give me the number that represents G in the formula.Pendragon5 (talk) 21:15, 5 February 2012 (UTC)[reply]

Why? It doesn't make a difference in this case. (4π²)r³/T² ≡ (4π²r³)/T² ≡ 4(π²r³)/T² ≡ 4π²(r³/T²) No point in putting in unneeded brackets, but then I always lean more maths than physics. The '/' as division is a programming convention, I think, which has crept into casual use where people are communicating with mostly plaintext over keyboards. You could use ÷ instead, if you prefer. 86.166.41.126 (talk) 21:22, 5 February 2012 (UTC)[reply]

There isn't a single number which is "G" - it's a physical quantity, which means it has both magnitude and units. What number you use depends on the units system you are operating under. If you're using standard SI units, then G ≈ 6.674 × 10^-11 N×(m²/kg²) . The equation as written is valid, however, for any linear unit system where zeros are appropriately placed. For example, you could use a G ≈ 3.730 × 10^-7 furlong³/(troy ounce×(fortnight²)), if those were the units you were measuring the other quantities in. -- 140.142.20.101 (talk) 22:03, 5 February 2012 (UTC)[reply]

I'm just like lost right there. I have no idea which unit i'm suppose to be in. Alright let just make up a problem. 2 stars with the total solar mass is 2, separation distance is 5 AU. What is the period? The answer unit is in second or minutes or hour? Which number is G in this case?Pendragon5 (talk) 22:50, 5 February 2012 (UTC)[reply]

Well, it's up to you what units you want. If you want seconds (easy to convert to minutes or hours), then the easiest way of writing G is 1.124 AU³ M_sun s^-2, which means that you can plug the numbers straight in. That said, it's almost certainly better to put the units in SI - 1 solar mass is ~2 x 10³⁰ kg, 1 AU is ~1.5 x 10¹¹ metres. Then you can use the SI version, G ≈ 6.674 × 10^-11 N×(m²/kg²) which is the version that you'd be given in most text books. Smurrayinchester 23:43, 5 February 2012 (UTC)[reply]

It is not just converting. I know how to convert and it's easy yea. OK let say i do that problem, at the end i will get a number right? That number will have a certain unit in second, minute, hour? It can not be randomly one of them then you can convert it, it doesn't work that way. "Which means that you can plug the numbers straight in"? I don't get this? I don't understand all of this AU³ M_sun s^-2 after the number 1.124. So if i plug the number in my calculator, what number is G? This is just way too confusing, i don't really get any of the explanations.

OK how about just do one problem as an example for me. It's better than just explaining around. 2 stars with the total mass of 2 solar mass, separation distance is 5 AU. What is the period? <---- answer this question and show me how you did it. Write out the equations with numbers that can be plug in the calculator.Pendragon5 (talk) 02:11, 6 February 2012 (UTC)[reply]

If you're looking for the period anyways, just do the algebra first to get yourself in a form of T= blah blah. If you look at Orbital_period#Two_bodies_orbiting_each_other it's already in that form. The article uses "P" instead of "T", but just plug the numbers in. If we assume a nearly circular common orbit between two bodies of exactly one solar mass each, of a distance of 1AU, use:

$P=2\pi {\sqrt {\frac {a^{3}}{G\left(M_{1}+M_{2}\right)}}}$
P = what you are looking for
π = 3.14159...
G = 6.674x10^-11 N m² kg ^-2
(m1 + m2) = 2x(1.9891x10³⁰) kg
a = 149.60x10⁹ m

Since the G constant uses SI units, then the time unit is seconds; because N = kg m s^-2, so the Newton unit already has seconds in it. If you do the dimensional analysis properly, you get units of P in seconds. So, plug those numbers in the equation, solve, and viola. You're done. --Jayron32 03:33, 6 February 2012 (UTC)[reply]

I think that should be a = 5 x 149.60x10⁹ m. Gandalf61 (talk) 08:57, 6 February 2012 (UTC)[reply]

It's not a complicated concept. Say you want to calculate the distance that light travels in time t. d=ct, where c is the speed of light. If you choose c to be in m/s (standard SI), you have to put t in seconds, and d will be in meters. If you choose c to be in cm/nanosecond, you have to put t in nanoseconds, and d will be in centimeters. --140.180.15.97 (talk) 08:12, 6 February 2012 (UTC)[reply]

In this problem, it is easier not to go via SI units. We know one orbit for which the equation is fulfilled, that is earth's orbit with

a=1\,\mathrm {AU}

,

M_{1}+M_{2}=M_{\mathrm {sun} }

(earth's mass can be safely neglected) and

P=1\,\mathrm {yr}

. These are convenient units for this sort of problem, especially since OP's data are given in those units. Observing that

P\propto a^{3/2}(M_{1}+M_{2})^{-1/2}

,

we can write the equation as

{\frac {P}{1\,\mathrm {yr} }}=\left({\frac {a}{1\,\mathrm {AU} }}\right)^{3/2}\,\left({\frac {M_{1}+M_{2}}{1\,M_{\mathrm {sun} }}}\right)^{-1/2}

With the numbers given by OP, the result is

P=5^{3/2}2^{-1/2}\,\mathrm {yr} =7.9\,\mathrm {yr}

. In those units (AU, yr, solar mass), the gravitational constant is simply

1/4\pi ^{2}

. --Wrongfilter (talk) 08:41, 6 February 2012 (UTC)[reply]

So the equation of Wrongfilter will result the answer in year, and the equation of Jayron will result in seconds, right?Pendragon5 (talk) 20:30, 6 February 2012 (UTC)[reply]

Yes, or you could take the answer from my section and divide it by 31,557,600 to get years, or you could take the answer from Wrongfilter's version and multiply it by 31,557,600 to get the answer in seconds. Or, knowing the conversions (1 minute = 60 seconds, 1 hour = 60 minutes, 1 day = 24 hours, 1 year = 365.25 days) you can take any of the answers and put it into any arbitrary time unit you like. --Jayron32 01:49, 7 February 2012 (UTC)[reply]

Center of gravity below the center of mass

Does anybody know how the location of Petronas Towers center of gravity was calculated in the "Young H. D., Freedman R. A., Sears and Zemansky University Physics"? There are different approaches/approximations in calculating CG position in non-uniform field. (Or, at least, how to find our discussion page on CG after the article was merged into the Center of mass article?)--Ilevanat (talk) 21:30, 5 February 2012 (UTC)[reply]

Talk:Center_of_gravity. StuRat (talk) 23:05, 5 February 2012 (UTC)[reply]

Thank you! I would not have found it myself. Still, does anybody know anything about the Sears and Zemansky textbook result?--Ilevanat (talk) 01:17, 6 February 2012 (UTC)[reply]

The variation in the force of gravity over the height of a building is so small that it can probably be ignored entirely. So, I certainly wouldn't worry about which formula they use to compensate for the non-uniform field, as any difference between them is sure to be insignificant. The CG would vary more with the movement of people through the building. StuRat (talk) 08:51, 7 February 2012 (UTC)[reply]

My simplistic model, which incorrectly assumes each floor of the Petronas Towers has the same mass, finds that the center of gravity is less than an inch from the center of mass. Someguy1221 (talk) 09:13, 7 February 2012 (UTC)[reply]

It indeed is, according to the textbook (about 2 cm). Surely, nothing significant from the practical point of view (and the observation by StuRat might be true in some sense). But the discussion we had about the former CG page was related to the conceptual and historical differences between CG and CM (and it is partly reflected in the present CM article, but I think this could still be improved). Anyway, Someguy1221, I will probably bother you some more soon at your talk page.--Ilevanat (talk) 02:05, 8 February 2012 (UTC)[reply]

lamp starters in series circuit fluorescent lamps e.g. philips lighting s2. how are these different from ordinary lamp starters e.g. philips lighting s10?

when two 20 watt fluorescent lamps were arranged in parallel each with its own lamp starter but series-connected to a single 40 watt choke only one or the other lamp lit up. arranged in series, each lamp parallelled with a philips lighting s2 starter on two lamp terminals and series-connected to the 40 watt choke both lamps lit up as desired.

what is different about an s2 lamp starter compared with the more ordinary lamp starters such as an s10 both manufactured by philips lighting?

thank you for your interest. — Preceding unsigned comment added by Davidlixenberg (talk • contribs) 21:35, 5 February 2012 (UTC)[reply]

If you make two flouro tubes in parallel share a common ballast, I would expect only one will light up, and even if both fire, only one will stay lit. This is because when a tube is not ionised (ie not lit up) it is an open circuit. This allows the ballast to resonate with the starter to produce a high voltage to close the starter and provide heater current to the tube. Due to manufacturing variation, one starter will close at a lower voltage - this will prevent the other starter from allowing heater current in its tube. Once one tube is going, it "clamps" the voltage well below that which will fire the its own starter and the other starter - so the other tube cannot start. Keit124.182.1.238 (talk) 02:41, 6 February 2012 (UTC)[reply]