Wikipedia:Reference desk/Archives/Science/2006 November 9

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November 9[edit]

Skid Marks[edit]

Is it possible to determine the amount of rubber removed from a cars tyres by measuring the length of the skid marks? Or does this depend upno the cars weight as well? Also, how do the police tell what speed the vehicle was doing just by the length?--Light current 00:35, 9 November 2006 (UTC)

I doubt the length is enough to tell the amount of rubber removed. Heavier cars with a softer tire would leave more rubber on the ground, I'm fairly sure. However, how much it varies by, I have no idea. The police can estimate how long it takes the vehicle to break by how long the skid mark is, which is related to the speed, but I'm not sure how accurate that is either, but cars generally have stopping distance statistics. --Wirbelwindヴィルヴェルヴィント (talk) 02:11, 9 November 2006 (UTC)
Well, you can estimate how much rubber there is per square centimetre of the skid mark, and then multiply this by the skid mark's area. --Bowlhover 03:17, 9 November 2006 (UTC)

THe problem as with other things is estimating how much rubber has been spread per sq cm of road. Would they have to scrape it off and weigh it. Or could a solvent be used to remove the mark then the liquid be analysed? Actually I dont think tyre rubber is soluble in anything--Light current 13:15, 9 November 2006 (UTC)

The stopping distance depends on the weight of the car in this case. (The stopping distance given by the manufacturer is usually the distance it takes to stop without skidding). Assuming that the driver slammed the brakes, thus causing the skid marks to appear the instant the decceleration started, the speed of the car can be found out quite accurately. Using the formula F=μN (μ is the co-effecient of friction between the road and the tyres. N is the weight of the car), the frictional force is calculated. Knowing the mass and the force, the decceleration can be calculated. Now we know the force that stopped the car, we know the decceleration and we know the distance S at which it stopped (That is the length of the skid marks). v=sqrt(2aS) will give us the velocity of the car just before braking. -- Wikicheng 04:53, 9 November 2006 (UTC)
Yeah I think youll find that the coefficient of friction varies wildly from start to finish of the skid marks--Light current 00:57, 10 November 2006 (UTC)
I don't think so. The coeffecient of friction is a constant for a pair of surfaces -- Wikicheng 10:07, 10 November 2006 (UTC)
You're assuming the two surfaces have no changing properties: As tires heat up the rubber gets softer and more likely to grip on some surfaces but more likely to smear on other surfaces. There is also the question that noone has yet raised of ABS which throws off your calculation of sliding friction because the tire is only sliding for brief periods. --Jmeden2000 15:35, 10 November 2006 (UTC)
Well... you are right regarding the changing properties...:-) -- Wikicheng 17:07, 10 November 2006 (UTC)
I've seen the police doing tests after a deadly accident in which one car was still in good driving condition. They slammed the brakes on the same road at different speeds and measured each time the length of the skid marks. I further heard that NapiSan is effective.  --LambiamTalk 10:27, 9 November 2006 (UTC)

Joules and Farads and Volts, oh my![edit]

What is the formula that shows how many joules there are based on Volts and Microfarads? I am wondering how many microfarads it will take at 35 or 50 volts to equal one joule. Thank you. Ilikefood 01:51, 9 November 2006 (UTC)

If you look at the page on volts, V = J / C, or J = V * C. --Wirbelwindヴィルヴェルヴィント (talk) 02:15, 9 November 2006 (UTC)
that page is wrong then. I'm not sure what they are getting at. P = VI , Volts * couloumbs/sec. Energy is the integral of Power through time. Therefore, Energy = qV (q is the total charge due to integrating I, electron Volt is also energy makes sense from equation). I = C dV/dt. Therefore, Energy = integral (CV dV) = 1/2 * CV2 . The Capacitor article seems to have it correct. The volt article looks very confusing. I think they confuse J (current) with Joules (Energy) --Tbeatty 04:40, 9 November 2006 (UTC)
Someone should edit it before I make myself look bad again =D --Wirbelwindヴィルヴェルヴィント (talk) 05:30, 9 November 2006 (UTC)
Volts = Joules per Coulomb. That formula has nothing to do with capacitance -24.84.54.54 05:40, 9 November 2006 (UTC)
Um, right. So the correct answer is Energy (in Joules) = V^2 * F, right? ^^; --Wirbelwindヴィルヴェルヴィント (talk) 05:56, 9 November 2006 (UTC)
That's right. But for convenient units, it's often expressed as kV^2 * uF (kilovolts squared times microFarads).
Atlant 17:22, 9 November 2006 (UTC)
See in Capacitor the section " Stored energy:" The energy stored in Joules is given by:
where V is the voltage across the capacitor. So solving for capacitance in Farads, =
Multiply by 1,000,000 to get microfarads. For example, for 20 volts, 1 Joule would require .005 Farad or 5,000 microfarad. For 100 volts, 1 Joule would require only .0002 F or 200υf. Edison 18:08, 9 November 2006 (UTC)
So you're saying that voltage squared times farads equals joules? Ilikefood 21:40, 9 November 2006 (UTC)
I would go with voltage squared times farads divided by 2. Edison 23:28, 10 November 2006 (UTC)

Absolute minimum density, part II[edit]

I'm not sure how to ask this question, exactly. But building on the earlier question that asked about the absolute minimum density of empty space, what part do photons play in the density of space? According to the article Photon, photons have 'invariant mass', but I'm not sure I understand what 'invariant mass' is. Empirically, space appears to be full of photons... observing a distant star, I'm seeing the photons emitted from it in the distant past. If I move a small amount to one side, I still see photons emitted from that star. That tells me no matter where I stand in view of that star, I will see photons emitted from it. So space is full of photons, from that star and every other star. So can space be said to be 'full of photons'? Yet, being massless, photons don't really fill anything up, yes? 192.168.1.1 9:59, 8 November 2006 (PST)

Although a photon viewed as an elementary particle is massless, it has energy E = hν, where h is Planck's constant and ν is the photon's frequency. Therefore it adds to the invariant mass of any system to which it belongs (such as a compartment of space through which it is moving) by the famous equation E = mc2, by which m = E/c2 = hν/c2. Since there are unsolved problems with the "energy balance" of the cosmos (see Dark energy), I'm sure physicists have estimated the contribution of photons to the energy density of space, but I don't know the result.  --LambiamTalk 10:16, 9 November 2006 (UTC)
It might help to know that the energy density associated with a photon gas is (someone check my proportionality constant?), which means that, for instance, the cosmic background radiation has an energy density of , or a density of . This is equivalent to one electron per 7.85 cubic meters; very little! Of course, there are stronger light sources: direct sunlight has an energy density of , or . But this is still tiny; the density of water is , and an entire Earth's volume of that sunlight would mass only . Obviously the mean density in a galaxy is somewhere between direct sunlight on earth and the CMB. --Tardis 04:03, 10 November 2006 (UTC)

IC Accelerometer[edit]

Ant idea on how the IC ADXL202 measures acceleration? —Preceding unsigned comment added by 210.212.194.210 (talkcontribs)

gooooooooogle Weregerbil 09:57, 9 November 2006 (UTC)
See [1]. A micromachined spring holds surfaces apart. Separation affects capacitance. F=MA. Distance, i.e. capacitance, is a function of acceleration. Edison 17:33, 9 November 2006 (UTC)

Step down Transformers[edit]

We use Iron core step down transformers,which suffer from heating and hysterisis loss. Why cant we use capacitors arranged in series instead to step down voltage more efficiently??—Preceding unsigned comment added by 210.212.194.210 (talkcontribs)

Capacitors change the power factor and also as not useful (offer a lot of impedance) at low frequencies, which is a loss. --Wikicheng 10:28, 9 November 2006 (UTC)
Actually its not a loss, its a mismatch. No power is lost in a capacitor.--Light current 13:43, 9 November 2006 (UTC)
Also capacitors do not provide safety isolation from the mains supply. Some completely isolated low power devices do however use 'capacitor droppers'--Light current 13:41, 9 November 2006 (UTC)
Transformers maintain about the same voltage "transformation" ratio for any load within their rated capacity. They also don't care too much about the reactance of the load.
By comparison, it would be hard to design either a purely inductive or purely capacitive voltage divider with the same flexible characteristics (to accommodate a variety of loads). As Light current pointed out, you could do it for a single, constant load though, you'd still have those safety isolation concerns.
More and more, people tend to use off-line switching power supplies that first rectify the mains power, then convert it to high-frequency AC which is then run through a very small, low-loss transformer and possibly converted back to DC. For example, nearly all low-voltage lighting run from mains power now operates this way.
Atlant 17:28, 9 November 2006 (UTC)
It is quite possible to step voltage up or down with capacitors, or with storage batteries. To step down voltage, charge them in series and discharge them in parallel. To step up voltage, charge them in parallel and discharge them in series. Some electric utilities in the late 19th centiry used this method with storage batteries to transmit DC at 1,000 volts to distant substations to back up local low voltage DC generation. A switch was used to change over from series to parallel. A switch could similarly be used with caps, but electronic switching is more efficient for continuous operation. Practical capacitors do contain resistance and have I^2 R losses. The caps in series would divide up DC voltage partly in proportion to their leakage resistance, so some might fail from overvoltage unless resistors were connected across each cap to force equal voltage division. Cost and complexity are likely the reason that transformers are used instead. The LM2798 [2] says it somehow uses a switched capacitor to step down DC voltage, perhaps in a more subtle fashion than I discussed.The LM3351 is a capacitive step up device.[3] Edison 17:47, 9 November 2006 (UTC)
Transformers generally reflect any load impedance to the primary. So reactive load ---> reactive load to primary supply. Not what the utilities want!--Light current 23:36, 10 November 2006 (UTC)

Polarizers[edit]

Will the efficiency of polarizers be affected in case a polarizer film is attached to a transperant glass/plastic plate or sandwiched between them?? —Preceding unsigned comment added by 210.212.194.210 (talkcontribs)

Not much.
Atlant 17:32, 9 November 2006 (UTC)
I would think that it would depend on the refraction index of the material, since polarization filters light from certain directions, the glass attached to the exposure side of the film might defract the light hitting the film, and depending on the angle, cause some of that light to be filtered out instead of allowed through. I'm not sure though. --Wirbelwindヴィルヴェルヴィント (talk) 18:41, 9 November 2006 (UTC)

LIVER AS METABOLIC BUFFER[edit]

It is said that liver acts as a metabolic buffer/How does it act so......can anybody tell in this detail...its not a home work question...when iam studying metabolism the thing i observed is that each and every metabolic cycle somewher touches liver...so i want to know about this..... --hima 12:06, 9 November 2006 (UTC)

Welcome to Wikipedia. You can easily look up this topic yourself. Please see liver. For future questions, try using the search box at the top left of the screen. It's much quicker, and you will probably find a clearer answer. If you still don't understand, add a further question below by clicking the "edit" button to the right of your question title. .--Shantavira 12:54, 9 November 2006 (UTC)

The liver takes up glucose when blood sugar is high, and put sugar into the blood when blood sugar is falling. In other words, in many ways, it can shift its metabolism and what it takes up and puts out in response to fed and fasting states to maintain metabolic homeostasis for the organism. alteripse 13:38, 9 November 2006 (UTC)

BLOOD[edit]

Is blood red or blue? As far as i know blood is red only because of the oxygen on the cells and when thats gone you turn blue....so what colour is it?

It is red when oxygenated in arteries and bluish when deoxygenated in veins. alteripse 13:35, 9 November 2006 (UTC)

Is it blue or is it more maroon (purple)? And when they take a blood sample does it matter that its taken from a vein?--Light current 13:39, 9 November 2006 (UTC)
Blood is never blue. Many textbooks have simplified diagrams with red arteries and blue veins, while blood in the real world is either bright red or darker red. --Wjbeaty 02:59, 10 November 2006 (UTC)
I think the taking from a vein matters more in the sense that taking blood from an artery is like trying to have a leisurely sip from a firehose. Confusing Manifestation 14:35, 9 November 2006 (UTC)
We routinely sample blood from arteries - it is not difficult in the least. The reason it is not more widespread is a) good candidates (such as the radial artery) are deeper and hurt more than veins and b) the risk for complications if a clot forms is higher. InvictaHOG 16:56, 9 November 2006 (UTC)
So the body has more veins than arteries and they are more accessible (nearer the surface). Is that right?--Light current 16:59, 9 November 2006 (UTC)
I actually think that there are more veins than arteries, but I don't think that I have ever seen that written anywhere. It's not that arteries are necessarily further from the surface. It's just that the ones you'd feel comfortable sticking are further away (ie the carotid artery is pretty easy to poke but it's kinda important!) InvictaHOG 20:25, 9 November 2006 (UTC)
There certainly are more veins than arteries. The arteries are (just about) always accompanied by at least one vein, but the superficial veins are not accompanied by arteries. 62.16.185.24 22:51, 9 November 2006 (UTC)
There's likely minimal difference in the number of arteries & veins--the closed system requires something close to a 1:1 ratio. But veins are almost always closer to the surface of the body than arteries. Often veins sit right on top of arteries. -- Scientizzle 17:09, 9 November 2006 (UTC)
As the danger from damage to an artery is much higher than from damage to a vein, it is no wonder that natural selection made the arteries lie further inside. To the orginal questioner: I hope you are not misled by the usual pictures in biology books that depict arteries in red and veins in blue. This is only to allow to distinguish them. Of course, the choice of colours is influenced by reality: oxygen-rich blood is bright red, and oxygen-poor blood has a darker, slighly purple-ish red. Simon A. 18:58, 9 November 2006 (UTC)
So why are subdermal bloodclots blue? Such as a black eye, which is really blue, but then also red, yellow and even greenish. Where do those colours come from? DirkvdM 09:55, 10 November 2006 (UTC)
The different colors are due to different decay products choleglobin/verdoglobin (black), biliverdin (green) and bilirubin (yellow/brown), according to the German hematoma entry. -- Rwst 15:15, 10 November 2006 (UTC)

Chemisty Question about lollipops[edit]

Why do lollipops turn sugary? You know, they start out as solids and then if you dont eat them, they turn sticky and gooey. Why?

Have you tried reading the answers to the exact same question further up this page? See #Chemistry: lollipops. Confusing Manifestation 14:37, 9 November 2006 (UTC)

optical[edit]

Describe the process of prism thinning.

see [[4]]--Light current 15:05, 9 November 2006 (UTC)

Body Sensor[edit]

What is one's body sensor? Have been unwell for 2+ weeks with middle ear virus and doctor believes I should have made more of a recovery by now. He is sending me for a scan as he thinks it is my 'body sensor'. What kind of scan will it be? MRI?

J Wheeler

Your balance sensor is located in the inner ear and comprises the system of 3 semicircular canals mounted mutually orthogonally (my favorite word). THese canals are filled with fluid that sloshes about when you move your head and tells your brain which way up your head is . To look at your middle ear it will probably be an MRI scan. Im not sure if that can do it with ultrsonics nowadays.--Light current 16:47, 9 November 2006 (UTC)

Crossing legs and weight[edit]

If one cannot cross ones legs comfortably, does that indidcate that one is overweight?--Light current 17:30, 9 November 2006 (UTC)

Male and female pelvis design also affects this.
Atlant 17:34, 9 November 2006 (UTC)
It indicates the person has a narrow waist/pelvis, thats why the position is usually more confortable for women but I guess fat legs doesn't help + guys crush their balls so even less comfy. Keria 18:00, 9 November 2006 (UTC)
Yeah well I didnt want to mention that one. Ouch!--Light current 18:04, 9 November 2006 (UTC)
Ouch? Does it hurt for you when you cross your legs? I don't think "crushing the balls" hurts, unless you crush them hard. --Bowlhover 03:09, 10 November 2006 (UTC)
You are obviously a woman--Light current 07:57, 10 November 2006 (UTC)
Yes, how did you know? Actually, I'm not. I'm a boy who loves to cross my legs. For me, it doesn't hurt at all (does it hurt for you?). --Bowlhover 02:36, 11 November 2006 (UTC)
Guys tend to cross their legs by resting the ankle across the opposite knee, thus avoiding ball/cock-compression. --Kurt Shaped Box 23:59, 9 November 2006 (UTC)
Im afraid Im not allowed to comment! 8-(--Light current 00:04, 10 November 2006 (UTC)
Well, let me then. Right now I sit with my legs crossed knee over knee, as I often do (even more than ankle over knee, I think). Never a problem. Maybe I let my balls hang loose more than others. :) Tight pants might be a problem. I prefer loose clothing. Maybe that's it. DirkvdM 10:02, 10 November 2006 (UTC)

Also I ve noticed small children rearely cross their legs and seem to have difficulty doing so. Is this because of the undeveloped pelvis or the higher propartion of body fat?--Light current 15:43, 10 November 2006 (UTC)

I've noticed they often sit like this (top view of head and legs), which requires a flexibility which most adults lack:
  /\   /\
\/  \ /  \/
     O
StuRat 22:00, 10 November 2006 (UTC)
Looks like the old symbol for a thermistor--Light current 22:03, 10 November 2006 (UTC)
Kids are generally more limber than adults. Robovski 04:46, 11 November 2006 (UTC)

Mammal kidneys[edit]

What mammal has kidneys that are at least four times more efficient at retaining water and excreting salt than those of humans?

It is a fairly broad question, but a reasonable place to start looking would be marine mammals or desert dwelling organisms like the kangaroo rat. --TeaDrinker 20:33, 9 November 2006 (UTC)
Camel ? StuRat 00:16, 10 November 2006 (UTC)
Dolphins or whales--Light current 02:00, 10 November 2006 (UTC)
(Sound of klaxon) Dolphins do not drink (according to QI). --Shantavira 08:13, 10 November 2006 (UTC)
Of course, if their food contains a great deal of salt, they must have some means of reducing the amount of salt in their plasma. If I recall correctly, most marine mammals have excretory output which has salinity above that of the surrounding water. Most marine mammals have a blood salt level which is significantly lower than the surrounding water, which indicates a kidney which is fairly efficient at removing salt. (Salt) water uptake by marine mammals is still fairly unknown; it has been demonstrated (again, by my recollection) for some species of seals. The thinking is that the water contains nutrients not found in the food... But this is a bit outside my official research, so a real expert may be able to correct me. (I recall reading that the Kangaroo rat can raise its urine salt content 14-17x the level in its plasma, that's why I included it...) --TeaDrinker 21:11, 10 November 2006 (UTC)

science[edit]

where can I go to find out about different subatomic particles and their charges?

Our List of particles might be of use to you. Let us know if we can help you further :-) --HappyCamper 20:32, 9 November 2006 (UTC)

Boiling-point elevation[edit]

Why does boiling-point elevation happen? A.Z. 20:34, 9 November 2006 (UTC)

A simple way to think of it is like this: take some liquid, and add heat. It boils if enough energy is put into overcoming the attraction forces between the molecules of the liquid. Now, if you add something that increases this interaction, then naturally, the boiling point will go up. Say, when you put some sugar into water. This is an overly simplified picture though. The mass of the particles, the size, the dipole moment, etc...all come into play. Actually, you might want to look at vapour pressure too. The effect of adding solute to solvent lowers the vapor pressure. Anything else the reference desk can help you with? --HappyCamper 20:42, 9 November 2006 (UTC)
There was an episode entitled "DEAD RECKONING" of "Science Fiction Theatre" first shown September 17, 1955, in which an airplane, maybe a DC-3, had a broken altimeter, and was unpressurized, and the pilot needed to fly above the mountains without going so high he would pass out from hypoxia. Fortunately there was a professor on board who had a hotplate, a pan of water, a thermometer, and a thorough knowledge of Boiling Point Elevation. By boiling the pan of water on the hotplate and observing the temperature of the boiling water, he could accurately determine the altitude, and get the plane safely through. If the plane climbed, the boiling point would decrease immediately; if it descended, the boiling point would immediately increase. Edison 05:50, 10 November 2006 (UTC)
Note that the variation of the boiling point with elevation is not the same as boiling-point elevation, though! --Anonymous, 00:02 UTC, November 11.
Isn´t there a situation in which adding a solute to a solvent will decrease the boiling point? Why should "The mass of the particles, the size, the dipole moment, etc..." of the solute always increase the intermolecular interactons of the compound?A.Z. 17:29, 11 November 2006 (UTC)

Diseases which warrant a Quarantine[edit]

I read on another website that there are nine human diseases for which a quarantine can legally be forced; the most common in the United States is Plague. All I ask is a list of the other eight, or if this number is incorrect, the correct list. I just want to know! Thank you very much!

Google is your friend: according to this web page, the diseases are: cholera, diphtheria, infectious tuberculosis, plague, smallpox, yellow fever, viral hemorrhagic fevers, and (since 2003) SARS. Simon A. 22:09, 9 November 2006 (UTC)
Just to clarify, for these diseases, isolation is warranted (but not mandated). Quarantine, which refers to the sequestration of individuals, plants, animals, or items which do not necesserily show symptoms but have been exposed or have a suspected capability to transmit infection etc., is much more broadly used, and in some cases is practiced with little to no explanation (I think with a 20 day limit). Some info (for one state) in the code books at 105 CMR, specifically section 300.200. Tuckerekcut 22:20, 9 November 2006 (UTC)

I should point out...that what can be "legally enforced" varies greatly between countries. Different countries have different laws regarding quarantine. --`/aksha 11:13, 10 November 2006 (UTC)

Capacitors[edit]

When you put a capacitor across a battery, it raises the battery's power, but how come when you apply something to that, the electricity doesnt just skip the capacitor? I thought that electricity always took the path of least resistance, so if there's a lightbulb or something connected to both of them, why does electricity still go to the capacitor and raise the voltage? For example:

         +                       -
   ___ /---Capacitor--\____
 /      \----Battery---/       \
 \_______           ___________/
              Light (or anything else)

if the capacitor is taken out, the light is dimmer. I don't understand how the electricity still goes through the cap. so could somebody please tell my whats going on? Thanks Ilikefood 21:47, 9 November 2006 (UTC)

Most, but not all, electricity takes the path of least resistance. StuRat 00:08, 10 November 2006 (UTC)
I don't understand; steady DC current doesn't flow through a capacitor. Melchoir 00:25, 10 November 2006 (UTC)
I think the key is that the voltage from the battery is not constant. The capacitor then acts sort of like a regulator. --HappyCamper 00:35, 10 November 2006 (UTC)
THe battery has a certain internal resistance. THis can be quite high if the battery is nearly discharged. When the capacitor is attached it will charge to the o/c voltage of the capacitor. If a lamp is then connected, the capacitor will supply most of the energy to the lamp until it discharges to the on load voltage of the battery. At this time, the energy will be supplied solely by the battery and since there is a resistance in series with the voltage source, the lamp will be dimmer.--Light current 01:03, 10 November 2006 (UTC)
It is absolutely false that electricity takes the path of least resistance. In fact, electricity takes ALL possible paths, and divides in inverse proportion to the resistance of each path (for DC) or the impedance of each path (for AC). Consider that in your home, the various devices are all connected in parallel, and each operates. If electricity took only the path of least resistance, then only the one lowest resistance device would operate at a time. Instead, the low wattage (high resistance) device draws a little current, while the high wattage (low resistance ) device draws a lot of current. A capacitor across a battery charges up to the battery voltage. When a load is connected across the battery and capacitor combination, the battery AND the capacitor discharge, increasing the initially available current because of the decrease in the internal resistance of the combination. The battery supplies current to the load, and the capacitor supplies current to the load. There is no "skipping." The output voltage of the capacitor will drop as it discharges. The output voltage of the battery will drop as it is discharged and chemical effects such as polarization (hydrogen bubbles etc) limit its output voltage. Edison 06:01, 10 November 2006 (UTC)

Number of Species in Asia[edit]

Does anyone know and/or have a source for a rough estimate of the amount of species of the following in Asia:

  • Fish
  • Mammals
  • Birds
  • Reptiles
  • Amphibians
  • Insects

Thank you. --24.247.126.44 23:22, 9 November 2006 (UTC)

List of Asian birds claims to be complete, but that's all I found. Melchoir 00:06, 10 November 2006 (UTC)

Another Curious Question for Anyone and Everyone to Answer[edit]

For the human eye in general, is it easier to detect and see a black dot on white, or a white dot on black? Hmm? Which would have less noise or more visibility? Or are they equally seen well? Thanks a bunch.

I remember reading somewhere that it is easier to see white text on black. Maybe this can help answer your question? --HappyCamper 00:37, 10 November 2006 (UTC)
Black on white is certainly easier for me to read. After a page of white on black, I can see after-images and my eyes go a bit squiffy for a few seconds. The absolute worst for me is red text on a dark blue background - the words actually seem to wobble around on the page. --Kurt Shaped Box 00:39, 10 November 2006 (UTC)
I can explain this last bit: red and blue light have sufficiently different wavelengths that the dispersion of your eyes' optics makes it impossible to focus on red and blue objects at the same distance simultaneously. When you have a close mix of red and blue like that, your eyes have a miserable time trying to focus on the page/screen. It's sometimes interesting to look at a mix of red/blue text on black; as you read, you can watch the words of the two colors alternate going in and out of focus. --Tardis 03:11, 10 November 2006 (UTC)
Blue is always difficult to focus on with a black background, regardless of other colors in the field. Colors with smaller wavelengths are not as good at illiciting pupillary constriction (for a variety of reasons), that's (one reason) why blue signs tend to look blurry at night (the backlit kind, that is). This is why not many cars use blue for gauges (not that I don't have the utmost respect for VW...). Blue gauges are both hard to focus on, and also tend to ruin your night-vision because so much light gets in past the wide open pupils. In contrast, orange and red lights constrict the pupil rapidly, preserving the rods and cones (which take longer to recover sensitivity than the pupil takes to dilate). As for the original question, it is easier to see a white dot on black, it is generally easier to detect a spot of signal in a sea of non-signal, in just about any sensory system. Although there are plenty of endogenous additions to the visual system, this still holds true. Really, it is not much different than asking, "is it easier to feel a lump in my mattress, or a clear spot in a field of lumps?" Tuckerekcut 04:39, 10 November 2006 (UTC)
I suppose that it is easier to recognise a white dot on a black surface than the other way round. Black means no light. A spot of light on a no light zone must be easier to spot than a tiny 'no light' zone on an 'all light' (white) zone ? In college physics labs, we are told to take the readings at minimum light rather tham maximum light in polarizastion experiments -- Wikicheng 10:16, 10 November 2006 (UTC)
(after edit conflict)Kurt, don't the after-images indicate that white on black has a greater subjective contrast? So it is really over-effective and grey on black would also do. At least, that makes some sense to me. DirkvdM 10:17, 10 November 2006 (UTC)
The question asks for a threshold of visibility task rather than say what makes for comfortable reading. As a thought experiment: assume that the black surface reflects a total of 1/100 as much light per unit area as the white dot does to the observer's eye. With no white dot, a black surface of a certain area under a certain illumination reflects a total of 1 unit of light to the observer and the scene looks black. Now add a small white dot which by itself reflects 1 unit of light. Now the total light reaching the observer is 2 units. The result is a doubling of the total light, or a 100% increase. Now imagine a white surface which reflects a total of 1000 units of light. A black dot is added and the total brightness decreases by 1/1000 or a .1% change in total brightness. From this, it sounds like in general it would be easier to detect a white dot on black than a black dot on white. Note that in signal detection or threshold measurement experiments, you have to make the task difficult enough the observer is uncertain and there are errors, or the observer would always be correct and the data would be meaningless. So white dot on black should be more detectable than black dot on white. Edison 23:44, 10 November 2006 (UTC)