Science desk
< March 1	<< Feb \| March \| Apr >>	March 3 >

Welcome to the Wikipedia Science Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

March 2

Analytical mechanics for Russian meteorite

how do the specialists calculate mass , speed and path of such meteorite?--Akbarmohammadzade (talk) 04:27, 2 March 2013 (UTC)[reply]

All they currently have is various videos of the event. By comparing the time stamps and locations of the meteor in each video, it should be possible to determine the speed. Unfortunately, many of the time stamps are off, complicating matters. Still, with enough videos, they can figure it out. The path is a bit simpler. Various vids from various angles allow them to fairly accurately establish the path, at least once it was bright enough to appear on videos. There's also a possible impact crater on a frozen lake. Working backward from these, they can extrapolate to find the location in space, but this gets less accurate the farther back they go. The mass is perhaps the hardest. They can infer the mass from the magnitude of the explosions and amount of light, but other factors besides mass also affect this, like the materials from which the object was composed. StuRat (talk) 05:02, 2 March 2013 (UTC)[reply]

Just for information of others reading this: This apparently refers to the 2013 Russian meteor event. — Sebastian 06:16, 2 March 2013 (UTC)[reply]

Here is the details. --PlanetEditor (talk) 06:30, 2 March 2013 (UTC)[reply]

Mass is definitely the most difficult parameter to estimate, that paper does not even make an attempt. NASA put it at 10,000 tons based on data from a worldwide network of infrasound sensors. The most common and straightforward method of estimating asteroid masses is from their absolute magnitude, which is, of course, impossible in this case unless some telescope has coincidentally captured a picture of it while still in orbit. Even then, it is an extremely inaccurate method because the albedo and composition of asteroids varies enormously and is rarely known with any accuracy. Spinning Spark 12:26, 2 March 2013 (UTC)[reply]

StuRat — they have more than the dashboard videos. The infrasound sensor network, designed to detect nuclear detonations, got a huge amount of data from it. They also have satellite information of various sorts (weather satellites, of course, military satellites, probably). --Mr.98 (talk) 19:27, 2 March 2013 (UTC)[reply]

I doubt if any of those sources would be as good as live camera footage. Satellites, for example, may only scan an area periodically, so would catch the trail after it had been blown around in the wind a bit. The time stamps might be more reliable, though. StuRat (talk) 16:46, 3 March 2013 (UTC)[reply]

They actually have pretty good satellite footage of the plume. Which you'd know if you'd bothered to look it up. --Mr.98 (talk) 02:53, 5 March 2013 (UTC)[reply]

The plume is what I'm talking about. That's not as good as having live video of the meteor itself, which is what the ground cameras provided. StuRat (talk) 03:13, 5 March 2013 (UTC)[reply]

EMP

Suppose a terrorist group sets off a small non-nuclear EMP generator with the following parameters: Total energy 56.25 kJ (very small indeed as such things go); average power 281.25 MW; pulse length 200 microseconds; capacitance 64.5 pF; inductance decreases linearly with time from 1 microhenry at t=0 to 2.5 nanohenry at t=200 microseconds (this is a simple CFG-type device with the main coil itself acting as the antenna); frequency increases as the square root of time from 20 MHz at t=0 to 400 MHz at t=200 microseconds (the "Valkyrie battlecry", as I call it :-) ). Also assume that they detonate the device right next to a standard step-down transformer that reduces the voltage from 34.5 kV to 240 V, and the device couples into both the primary and the secondary windings of this transformer. What's the radius in which the device will fry most plug-in electrical devices? 24.23.196.85 (talk) 05:53, 2 March 2013 (UTC)[reply]

I gather the intent is to induce a spike or surge in the transformer, thus "frying" electronic devices in the circuit, right? I haven't done the math, but I suspect the transformer would either handle the pulse (which is weak compared to a lightning strike) - or the pulse would be strong enough to trip it, or otherwise cause a shunt or power failure. ~:74.60.29.141 (talk) 06:45, 2 March 2013 (UTC):~[reply]

Right, this is precisely how an EMP generator works. So, I gather that 56 kJ is too small for any serious damage to the transformer or anything else connected to it? 24.23.196.85 (talk) 07:19, 2 March 2013 (UTC)[reply]

At the timings nominated, the transformer should be modelled as a transmission line, however, the standard lumped model will give us a ball park idea. Unfortunately you have not nominated the size of the transformer, so I'll use the smallest (and most sensitive) size commonly used in street distribution: 10 kVA (per phase load 14 A). This can be expected to have a secondary side leakage inductance of ~~1 mH. This considerably exceeds your EMP equiv inductance, so we can simply take direct connection of the capacitance to get worst case voltage. With 200 pF capacitance, the peak voltage from 56 kJ is 30 MV. With a 200 uSec discharge time, this will certainly damage the transformer insulation. So we do need to calculate the range at which transformer breakdown will occur.

Now for a bit of order of magnitude estimation that I pored scorn on in an earlier question. In this case this is all we can do as you have not supplied a lot of critical information. The initial electrostatic field will be ~~30 MV/m (from 1 uH initial inductance) decreasing with the cube of the distance in meters. Thus, to get within the transformer's secondary insulation voltage breakdown limit, which I'll assume to be 6 kV, the maximum range to ruin the transformer will be ~~ D_{wireseparation} x (V_init/V_trans)^1/3 ~~ 6 m with typical wire separation in open wire systems. For underground distribution, it would thus seem that 74.60.29.141 is correct, it is not sufficient to destroy the transformer. This estimate can only be a very very rough guide.

Ratbone 124.182.176.152 (talk) 10:38, 2 March 2013 (UTC)[reply]

Re: "This will certainly damage the transformer insulation", try modeling it again with lightning arresters at the input to the transformer and at the service entrance to the house. (And possibly at the plug strip the equipment is plugged in to.) You might also want to model the fact that "assume that they detonate the device right next to a standard step-down transformer" does not mean that 100% of the energy will couple into the transformer. There will be losses because some energy goes in other directions, losses because of reflections caused by impedance mismatch, and -- this is the big one -- losses because the transformer is surrounded by a grounded faraday cage. --Guy Macon (talk) 12:39, 2 March 2013 (UTC)[reply]

If you read my post more carefully, it was in two parts: the first part was to see if damaging the transformer was at all possible - calculation with a direct connection is the easiest to model. This showed that destruction is possible, and so further work is required. In the second part I did that further work and established that the EMP device needs to be within roughly 6 m, at which range the coupling is indeed well below 100% - about 0.02% in fact. The fact that the transformer housing can act as a faraday cage is not relavent - I assumed all coupling was via the open wire distribution, allowing a coupling factor for the wire spacing. Due to series inductance in the distribution wiring, which in practice will be very large compared to the flashed over transformer, not much energy will go down the distribution into customer's equipment, and any surge/lightning protection at customer premises is not relevent as far as the transformer goes. It is normal practice in the small distribution transformers I assumed for calculation purposes not to have lightning/surge protection fitted. In any case, such devices are intended to shunt lightning to earth (where it wants to go), which has absolutely no relavence to any neaby EMP device (which has no need to go to earth). Like other posters, as far as larger transformers are concerned, and underground plant, I don't think such an EMP device would cause any damage. However, I have established that for small transformers in open wire installations, damage is possible with the EMP device a few meters away. Ratbone 120.145.182.161 (talk) 14:44, 2 March 2013 (UTC)[reply]

It certainly doesn't sound like it would have much effect to me. I'd have thought there would be far more efficient ways to destroy a transformer with the explosives you need for such an EMP weapon, and the surges on the power lines would probably not get far with the various devices to stop lightning strikes having an effect. Why would one waste it on such an enterprise when there are far simpler and better targets around? Dmcq (talk) 13:10, 2 March 2013 (UTC)[reply]

That it would be easier to just blow the transformer up with standard explosives is blindingly obvious. Any half witt terroist or local red-neck nutcase can make a decent bang with ordianry explosives. But EMP devices are high-tech advanced devices that will cost a heck of a lot more. Ratbone 120.145.182.161 (talk) 15:00, 2 March 2013 (UTC)[reply]

You might be interested in Report of the Commission to Assess the Threat to the United States from Electromagnetic Pulse (EMP) Attack. They do not highlight transformer insulation breakdown as being a particular problem. For large transmission transformers the greatest threat comes from damage to protective relays and other protective devices [1] for E1 pulses (the shortest duration and most relevant to this question). For small distribution transformers they point to arcing across the power line insulators possibly causing transformer explosion [2]. As for frying electronic equipment, they seem far more concerned with the direct effects of EMP on electronics rather than induced currents in the supply. For a larger (nuclear) EMP than described here, longer lasting ground induced currents could be a problem from the E3 pulse. Spinning Spark 16:05, 2 March 2013 (UTC)[reply]

Thanks, everyone! So, the bottom line is that the EMP attack will cause little or no damage in this case, at most burning out the transformer, right? In that case, here's a second scenario: the bad guys set off a similar device, but this time the total energy is 100 MJ (the highest energy practicable for a man-portable device) and the average power is correspondingly 500 GW, while all the other parameters remain the same. (Also assume that in this case the 34.5 kV power line supplying power to the targeted transformer in turn draws power from a 115-kV line via a substation just a few blocks away, which in turn delivers power from SoCalEd's Rinaldi Street substation that receives power directly from Path 26 -- a real worst-case scenario!) How much mayhem will happen in this second case? 24.23.196.85 (talk) 20:42, 2 March 2013 (UTC)[reply]

Just curious: is this background for a novel, or should you expect a ~~2:am knock on your door~~ [friendly visit] from the San Jose branch of Homeland Security? ~:74.60.29.141 (talk) 21:26, 2 March 2013 (UTC):~[reply]

Not to worry, this is for a detective novel. In fact, I don't think I could build such a device myself if I wanted to. 24.23.196.85 (talk) 21:47, 2 March 2013 (UTC)[reply]

And in any case, I'm an American patriot, so the LAST thing I would want to do is to cause widespread indiscriminate destruction to my own country's infrastructure! 24.23.196.85 (talk) 21:54, 2 March 2013 (UTC)[reply]

I would like to address the claim that "such devices are intended to shunt lightning to earth (where it wants to go), which has absolutely no relavence to any neaby EMP device (which has no need to go to earth)".

To keep things simple, I will assume a 1kV single phase two-conductor live with no connection to earth ground.

Assume that your EMP device can cause that 1kV line to jump to 100kV between the two conductors.

Now add a lightning arrester from each line to earth ground that limits the voltage to 2kV. Those lightning arresters would also limit the voltage between the two conductors to 4kV.

In real life there would usually be a third lightning arrester that would limit the line-to-line voltage to 2kV. --Guy Macon (talk) 22:37, 2 March 2013 (UTC)[reply]

But would it have time to activate (the whole pulse is only 200 microseconds long, remember)? 24.23.196.85 (talk) 22:57, 2 March 2013 (UTC)[reply]

[e/c] You might want to consider: line filters, surge protectors and Uninterruptible Power Supply; and check out: The EMP threat: fact, fiction, and response and Section 7 in this PDF which has info specific to your inquiry. ~E:[last modified]74.60.29.141 (talk) 23:59, 2 March 2013 (UTC)[reply]

There is a classic joke that goes something like this:

Two men see, in the distance, an angry bear heading towards them. One of the men begins to put on his running shoes. The second man says, "What are you doing? Are you crazy? Bears can run at 30 mph! Your sneakers won’t help you run faster than that bear!" To which the first man replies, "I don’t have to run faster than the bear, I only have to run faster than you."

Your protection circuit does not have to withstand the full voltage of an EMP or lightning strike. It only has to withstand the maximum voltage beyond which an arc forms between the conductors of your wiring. Your protection circuit does not have to be faster than the risetime of an EMP or lightning strike. It only has to be faster than the maximum risetime that the capacitance and inductance of your wiring allows. --Guy Macon (talk) 14:43, 3 March 2013 (UTC)[reply]

That is not right. Any circuit consisting of lumped or distributed inductance, capacitance, and resistance has a response amplitude directly proportional to the excitation voltage. This means that while the risetime, measured from (say) 10% to 90% of final amplitude, is determined by the inductance, capacitance, and resistance, the rate of change is not. The rate of change is directly proportional to excitation amplitude. Therefore, the time taken to reach a given (breakdown eg) voltage is inversely proportional to excitation amplitude. The bigger the EMP, the faster the time to reach breakdown. Ratbone 124.182.147.20 (talk) 03:49, 4 March 2013 (UTC)[reply]

(un-indent) Thanks for the info, everyone. So I gather (especially from the documents provided by 74 IP) that the main danger is to sensitive electronics (computers, cell phones, etc.), while the heavier-duty components that actually handle the power distribution (power-line wires, switchgear, circuit breakers, etc.) will remain intact? I guess then that I'll have to revise quite a number of details: for example, the air won't be "thick with the acrid stench of burning insulation" like I thought it would be; when one of the detectives goes to check the fuse box, he won't see "a mess of half-molten copper" sitting at the bottom; and the transformers at the Orange County Airport substation probably won't be burning all night long like a bonfire. On the other hand, the detectives' cell phones will indeed get fried and might even suffer thermal damage up to and including partial melting of the outer casing (especially if turned on while recharging) -- and it goes without saying that all power to the affected area will get knocked out for up to several days. 24.23.196.85 (talk) 05:54, 4 March 2013 (UTC)[reply]

Where did you get all that nonsense from? Let's take it one item at a time:-

Effect on home electronics, computers, and the like: Not likely to be affected, as they are isolated from an EMP device to your specification near a distribution transformer by the inductance of the street distribution wires. Nothing supplied by 74 IP refutes this.
Effect on cell phones: Quite unlikely, as the circuit board and inbuit antenna is simply not long enough to intercept sufficient EMP field. However, if the EMP device is set off adjacent to a cell phone base station (the little hut next to a cell tower containing the electronics), the cell base could be knocked out by damaging it's power supply or the land line connection to the phone company. But bear in mind they have battery backup, and the loss of one base does not prevent phones from working - it only reduces service.
Effect on power company distribution transformers: If the transformer is a small one, and the street distribution is not underground cabling, then an EMP device within a few meters will damage it, as I shoed by simple calculation.
Will any smells or molten copper be noticed? I doubt it.
Would power to affected area be knocked out for several days? Not likely, unless perhaps you are talking about a very backward area in a third world country. The most likey outcome is just a damaged transformer. Most power companies will detect (and in any case customers will ring them up when power fails) and replace a distribution transformer within less than an hour. Only if damage is widespread will they take longer, and even then you are only looling at a few hours outage at most.
Note that for the larger distribution transformers, and substations generally, and transformers of any size in an underground network, no damage will acrue from the EMP device you specified.
You should also note that in the case of significant public infrastructure, such as feeds to hospitals, airports, public utilities, electric railway supplies, and major substations, knocking out one substation will have at most only momentary (eye-blink time) effect on power available to customers, as redundant infrastructure is provided. For example, in my city, the hospitals, the telephone exchanges, the police headquarters, and certain other important buildings, are on the "priority ring" - this is a duplicated high voltage ring fed from 4 major substations. At least two substations must be knocked (on both busses) out to kill power on these rings. No single fault on either ring can kill power on that ring. Two simulataneous faults on a ring can at worst only cut power between the two faults, and the customers still have power from the other ring. You can expect the power company to attend to any priority ring faults immediately, not hours, not days, but immediately. Apart from satisfying service garantees, quite a lot of revenue, far outstripping the technician's wages, will be lost if a second fault occurs and major customers don't have power. Outside the city, important things like the international airport, hospitals, etc, are feed by two sets of cables from two geographically diverse substations.

Ratbone 58.170.164.191 (talk) 09:54, 4 March 2013 (UTC)[reply]

Force carrier particles in a black hole

If the fundamental forces of nature are carried in particles/waves, as quantum physics maintains that they are, then how are gravitrons (for mass) and charge carriers able to exert influence on the outside world, seeing as they cannot escape the black hole? Thus, a black hole should not be able to exert the full amount of force due from its mass nor its charge. Magog the Ogre (t • c) 15:45, 2 March 2013 (UTC)[reply]

A (theoretical) graviton is itself massless, so is not affected by gravity. How can it "carry mass" yet itself be massless ? Think of it as carrying information, like if you carry a bankbook with your bank balance written on it, but that doesn't mean you have the actual cash on you. StuRat (talk) 16:38, 2 March 2013 (UTC)[reply]

StuRat, sometimes the answers you give are very useful, but I wish you would stop answering questions where you don't have the slightest idea what you are talking about. Massless particles are indeed affected by gravity, for example photons are redshifted when they pass through a gravitational field. Regarding Magog's question, there is no accepted theory of quantum gravity currently, but if one is ever developed, it will probably have the gravitons created in the region outside the event horizon. The story is clearer for charge carriers. Quantum field theory allows for the creation of an electron-positron pair out of the vacuum, with one of them falling into the black hole and the other escaping to infinity -- this mechanism allows electric field propagation without anything actually escaping across the event horizon. Looie496 (talk) 16:53, 2 March 2013 (UTC)[reply]

To clarify, you mean particles with no rest mass can be affected by gravity, due to the mass they gain due to relativity effects at high speeds, right ? Yes, I neglected to include that (and our article on gravitons really should say it has 0 "rest mass", not just 0 "mass", but I can't figure out how to change it). StuRat (talk) 17:26, 2 March 2013 (UTC)[reply]

They don't gain mass, but they are affected by gravity because space time is curved. Light has a mass of exactly 0, but it cannot escape a black hole due to the curving of space time. Magog the Ogre (t • c) 17:38, 2 March 2013 (UTC)[reply]

But, again, photons have zero rest mass, but do have relativistic mass. StuRat (talk) 20:55, 2 March 2013 (UTC)[reply]

Gravity affects everything equally (equivalence principle); the mass really doesn't matter, whether it's rest mass or relativistic mass. Also, relativistic mass has never been a very useful concept, and isn't usually taught any more. It leads people to think that rapidly moving objects should collapse into black holes, for example, which they certainly don't. -- BenRG (talk) 23:30, 2 March 2013 (UTC)[reply]

So, in theory then, a black hole could have a hugely positive charge, but only the content which lay on the event horizon would be able to affect the outside world in terms of charge? (I realize that incoming material does not fall into a black hole but forever stays on the event horizon... but what about material that was present when the black hole expanded, or material that was at the center of the original star, say a magnetar, when the black hole formed) And what does this say about the speed of gravity? Magog the Ogre (t • c) 17:08, 2 March 2013 (UTC)[reply]

Matter does cross the event horizon. You can't see it cross from outside, because no light from the crossing can ever reach you, but it does cross. The electromagnetic field of a black hole has the same strength as the total field of the matter than went into the black hole, but it doesn't come from inside; it is a "fossil field" (retarded potential) from before the matter crossed the event horizon. The same is true of the gravitational field. This doesn't change (as far as anyone knows) in quantum mechanics; nothing that happens inside the event horizon affects anything outside. As for the virtual particle picture, I think this is just an example of a case where it isn't very helpful (as Count Iblis sort of says below). -- BenRG (talk) 23:30, 2 March 2013 (UTC)[reply]

Virtual particles are just mathematical tools to do computations, they are unphysical and don't stick to the equations of motions of real particles. E.g. in case of Compton scattering, two Feynman diagrams are needed to compute the amplitude. In one of these diagrams, the emitted photon leaves the electron before the photon that is going to be absorbed has arrived. Count Iblis (talk) 18:45, 2 March 2013 (UTC) che soale mahi .....[reply]

LCD panel colours from the side

Why do LCD panels look very red, green or blue from the side? Clover345 (talk) 17:03, 2 March 2013 (UTC)[reply]

To clarify, you aren't asking about those LCD monitors with "ambient display" (that is, they intentionally shine lights out the sides), right ? StuRat (talk) 17:38, 2 March 2013 (UTC)[reply]

No, I mean LCD TVs, PC monitors, phones etc. when you look at them side on or angled on dark screens, most seem to change colour to red, blue or green depending on the display. Clover345 (talk) 17:48, 2 March 2013 (UTC)[reply]

LCDs, in particular, seem to only provide a good image over a limited angle (both up-down and right-left). Apparently some of the colors can be viewed over a somewhat wider angle than others, giving the color mismatch you describe at the fringe. I'm not sure why this is, though. Could it be a prism effect ? StuRat (talk) 17:51, 2 March 2013 (UTC)[reply]

The way I understand it (no ref, may not be accurate...) is that the LCD mask over the subpixels only look right when viewed from the right angles. Otherwise, they mask out adjacent subpixels, switching which color they seem to block from your viewpoint. 38.111.64.107 (talk) 13:23, 4 March 2013 (UTC)[reply]

how practical are diy 3d laser scanners for specific industrial applicaitons?

I'd like to understand whether I could expect to be able to cheaply hand-assemble a relatively real-time 3D recognition engine with cheap DIY parts. (Relatively real-time meaning I don't know 200-500 ms delay or whatever. Cheap means under $100.) My problem though is that I don't know the fundamentals of the different possible 3D recognition approaches!!!

I know of several:

-> stereoscopy, meaning two cameras a known distance apart. you match pixels and then figure out how far you had to move them compared with the known difference between your cameras

-> then I know there is "structured light" meaning you shine some light like a straight line or something, and then from how it deforms, you can tell what the shape of the object must have been that it hit

-> there is even 3D from motion processing! (parallex).

how practical are any of these in a diy 3d project? Thanks!! 86.101.32.82 (talk) 19:27, 2 March 2013 (UTC)[reply]

I think we'll need a more detailed description of the application before we can help. As described, determining the position of the user's fingers on a table is a 2D rather than a 3D problem, and could be done with a single camera; or, indeed, you could use a touchpad and not require any optical sensors at all. Why do you need 3D? Tevildo (talk) 21:13, 2 March 2013 (UTC)[reply]

The application is of a resolution like trying to recognize the relative positions of hot dogs on a conveyor belt in 3D (so you can represent it as some cylinders in a coordinate system - this is about recognizing the xyz values of the ends of those cylinders) -- it doesn't really matter as this is a generic DIY project. this is just to indicate the level of granularity: i.e. we are not scanning a statue in mm resolution or anything like that, nor is it about a landscape or room or something like that. Basically, I am more interested in what kind of results I could get from this kind of DIY project and, importantly, what the best approach is. I'm unclear about even basic pros/cons, such as whether polyscopy (more cameras) would help, whether a greater distance between cameras would give me a better result trying stereoscopy, and so forth. Basically, I'd like basic references geared toward the practicalities I've mentioned. Thanks! 86.101.32.82 (talk) 23:29, 2 March 2013 (UTC)[reply]

See 3D scanner, Laser scanning, and Structured-light 3D scanner for our articles on the subject. A quick Google search on "laser scanner" comes up with lots of out-of-the box solutions at around the $300 mark, so a DIY system for $100 might be feasible. Tevildo (talk) 23:42, 2 March 2013 (UTC)[reply]

This is a fairly common conveyer belt application and the usual, and much simpler, solution is electronic detection using either a radar technique or distortion of an AC magnetic field. Since hot dogs have both a moisture content and ingredients that have a dielectric constant very different to air, detection is simple and reliable. Ratbone 124.178.153.245 (talk) 02:05, 3 March 2013 (UTC)[reply]

that's just the thing though

http://en.wikipedia.org/wiki/Laser_scanning
http://en.wikipedia.org/wiki/Structured-light_3D_scanner
http://www.google.com/search?q=stereoscopy+depth+recognition (where is our specific article)?
the just mentioned: http://en.wikipedia.org/wiki/3D_radar
http://en.wikipedia.org/wiki/Structure_from_motion

There are so many techniques!! I would like references, if possible, to help me compare them on a practical level. I realize there are "many ways of doing this" - so, please, what are the pros and cons? (In a mobile device that can detect fingers and the like). 86.101.32.82 (talk) 09:54, 3 March 2013 (UTC)[reply]

This site might be a better place to ask, although I would suspect their first question will be "What specific application do you have in mind?" (See this thread on the same forum for an example). Tevildo (talk) 16:11, 3 March 2013 (UTC)[reply]

The Kinect is an interesting 3D scanning device. It projects an IR pattern over the area and images it, along with taking color images, and is designed for gaming so it is real-time. Microsoft has a free SDK available for developing PC software that uses it. It isn't laser scanning, and I don't know if it is right for your application, but it seems a great entry-level way to get real-time 3d imaging. 38.111.64.107 (talk) 13:09, 4 March 2013 (UTC)[reply]

The problem with the Kinect in some of these applications is that the IR emitter and the camera that goes with it are quite far apart. This produces good results for large objects at long ranges - but starts to cause real difficulties for small objects at close ranges. The result is generally that you have to keep even small objects quite a distance from the Kinect - and then the poor spatial precision starts to become an issue. A new version of the Kinect is due to appear soon, it's claimed to be a big improvement in this specific situation, but it remains to be seen how good it really is. SteveBaker (talk) 14:37, 4 March 2013 (UTC)[reply]

For the literal case of hotdogs on a conveyor belt, I'd get the area as dark as you can and generate a laser line (shining a laser pointer into an appropriate lens works well - I used the guts of a discarded hand-held bar-code reader) pointing down onto the conveyor from vertically above. Use a regular digital camera to look at that line from a few inches away from the laser pointer along the length of the conveyor. What it sees is a straight line when there is nothing on the belt and a "cross-section" of the 3D shape when something intersects that line. If you know how fast the conveyor is moving, you'll get a bunch of cross-sections at known separations - each of which will have the shape of half of an ellipse. From that you can compute approximate elliptical cross-sections for each object as it crosses the laser line. The ratio of the diameters of those semi-ellipses gives you a hint as to the orientation of the object on the belt - which you can strengthen by stitching together multiple cross-sections over time and averaging to get the true orientation of the cylinder. It should be easily possible to do this in realtime - even at fairly fast belt speeds.

Total hardware cost should be around $40. Lots of software effort needed though.

SteveBaker (talk) 14:37, 4 March 2013 (UTC)[reply]

Pump & water question

Say you have a water pump with two independent pipes of the same diameter coming off of it. The pipes go up about 50 feet and each has a valve at the end. Pressure at the pump is 37 psi for both pipes, and 22 psi at the their ends. Opening the valve for one pipe changes the pressure in the second pipe by a very little - say 0.5 psi.

If you were to join these two pipes with a "T" fitting (same internal diameter as the 2 pipes), and read the pressure coming out of the "T", would you expect it to read 22 psi, 37 psi, or something in between?

(No this isn't a homework question - I'm hoping to come up with a clever solution for a real plumbing problem.)

Thanks! 142.68.14.116 (talk) 20:43, 2 March 2013 (UTC)[reply]

It will be somewhere in between, decreasing linearly with height. Are you trying to improve the plumbing in a high-rise condo? 24.23.196.85 (talk) 20:47, 2 March 2013 (UTC)[reply]

Why not tell us what the real plumbing problem is ? StuRat (talk) 20:50, 2 March 2013 (UTC)[reply]

Similar - trying to make a toilet flush on the bridge of a ship. I'm guessing the linear component would be .433 psi/foot due to gravity, plus some small friction-sort of losses? The toilet requires 35psi of water pressure for optimal operation, it's getting 22. 142.68.14.116 (talk) 20:53, 2 March 2013 (UTC)[reply]

Simplest solution, add a tank above the toilet. This approach is used for most land toilets, as it doesn't require much water pressure. Other than that, you'd need another pump, either replacing the current one with a more powerful pump, or adding a supplementary pump. Either sounds more complex than just adding a tank, to me. A tank will make the ship somewhat more top-heavy, but I can't imagine this tiny difference being significant. StuRat (talk) 20:57, 2 March 2013 (UTC)[reply]

StuRat, we're talking about this kind of tank, not that one. ;-) 24.23.196.85 (talk) 21:13, 2 March 2013 (UTC)[reply]

Tanks for clarification. :-) StuRat (talk) 02:08, 4 March 2013 (UTC) [reply]

With similar problems involving equalizing pressure differentials in racing engines (oil/fuel), the solution involves the use of pressure regulators (adjustable or static). (I.e.: decrease pressure to one fork of 'T' to increase pressure on other) ~E:74.60.29.141 (talk) 21:45, 2 March 2013 (UTC)[reply]

The only way I can see that working is to close off one of the pipes, sending all of it's water pressure into the other pipe. That should increase pressure somewhat, but I doubt if it would go from 22 to 37 PSI. This is basically the same as reducing the diameter of a single pipe to increase pressure.StuRat (talk) 05:28, 3 March 2013 (UTC)[reply]

Also, we should clarify. Atmospheric pressure is around 15 PSI, so, when you say 22 PSI and 37 PSI, do you mean that much pressure beyond atmospheric pressure (so 37 and 53 PSI, total) ?

Bernoulli's equation is easy, reliable and covers all questions and solutions on pressure in pipes and more. --Kharon (talk) 13:18, 3 March 2013 (UTC)[reply]

Considering that this question is about pressure drop along pipes and T-junctions, where the flow rate is a non-linear non-monotonic function of pressure drop, and a decidedly non-linear function of pipe diameter (4th power for stream flow; 19/7-th power for turbulent flow) depending on the Reynolds Number, please explain how Bernoulli's equation, which models the pressure as a linear function of flow rate, can be used to cover this question. Please explain how Bernoilli can explain the comparitively very high pressure drop in T-junctions and 90 degree pipe bends. Ratbone 124.182.147.20 (talk) 03:32, 4 March 2013 (UTC)[reply]