Wikipedia:Reference desk/Archives/Science/2010 September 26

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September 26

dihydroxy dimethylol ethylene urea

whats dihydroxy dimethylol ethylene urea —Preceding unsigned comment added by Kj650 (talk • contribs) 01:57, 26 September 2010 (UTC)

Its a fabric treating agent commonly abbreviated DMDHEU, and the formal IUPAC name is 1,2-Dimethylol-4,5-dihydroxyethyleneurea We don't currently have an article on that, but if you paste that name into Google you will find structures of it. It is used to create so-called "wrinkle-free" fabrics. See Permanent press which has some linked references you could follow for additional information. --Jayron32 02:11, 26 September 2010 (UTC)

how is it made —Preceding unsigned comment added by Kj650 (talk • contribs) 02:49, 26 September 2010 (UTC)

[1] this said it is made from urea, formaldehyde, and glyoxal. And for me this looks all right, although you have to be careful not to get a polymer as a end product.--Stone (talk) 07:18, 26 September 2010 (UTC)

Ford Escape Securilock keys

Just got 2005 Escape today. Found out after getting home that one key works on the locks and ignition, and the other key works on the locks but not the ignition. Am going back to the dealer tomorrow, but since it's closed right now and this question of what's the matter is driving me crazy, is it normal for one key to have lock-only privileges and is there any way for the owner to fix this himself? Thanks. 76.27.175.80 (talk) 02:12, 26 September 2010 (UTC)

Normal keys have three components: the physical cut of the key to fit the physical lock, the "remote" circuitry (lock and unlock buttons), and the transponder chip that is queried by the on-board computer before starting is permitted. Does your faulty key not fit the lock, or does it just fail to start the vehicle? If the latter, then the transponder has not been correctly programmed. You might be interested in this method of programming, but it will not work in your case because you have only one correctly-programmed key, so you need to take the key back to the dealer to be programmed by their specialist equipment. Dbfirs 03:37, 26 September 2010 (UTC)

Shelf life *after opening* Baileys Irish Cream?

The article for Baileys Irish Cream states that According to the manufacturer, Baileys has a shelf life of 30 months. It should be stored between 5 and 25 degrees Celsius, or 41 to 77 degrees Fahrenheit. Presumably this is for an unopened bottle. How long is Baileys safe to drink after being first opened? And should it be refrigerated? Doing so would keep it cooler than the manufacturer's recommended temperature... The Masked Booby (talk) 06:53, 26 September 2010 (UTC)

From their website:

Baileys Irish cream is the only cream liqueur that guarantees its taste for two years from the day it was made, opened or unopened, stored in the fridge or not when stored away from direct sunlight at a temperature range of 0-25 degrees centigrade. One of the keys to achieving this two year shelf-life is in our patented process of blending of fresh Irish cream with the spirits and the whiskey without the use of preservatives. The alcohol acts as a natural preservative for the product. Under normal conditions of storage Baileys Irish cream has a shelf-life of 30 months. If you are concerned about a bottle of Baileys Irish cream please check the best consumed before date on the bottle - all bottles now carry a best before date. This number is located on the bottom left hand side of the back label. Example : Code 11 20XY would mean that we guarantee the product would taste perfect until November20XY (XY is the year two years from the date of manufacture).

Like most products of this kind, I suspect the shelf life calculation has nothing to do with safety and is rather a statement that the taste might change over time. As discussed in shelf life, the terms "best before" and "best by" are generally suggestions based on the expectation that the items' quality and desirability will degrade over time, but are not an expression that the item is unsafe. By contrast, "expiration dates" and "use by" dates are specifically meant to indicate that an item may become unsafe after a period of time (or in the case of medicines, that it lacks the expected potency). Dragons flight (talk) 08:02, 26 September 2010 (UTC)

All I know is that my parents had an opened (but tightly capped) bottle of Irish Cream sitting on a shelf for God knows how long; and when I opened it, the contents had turned to an unpourable gelatinous mass. Deor (talk) 12:48, 26 September 2010 (UTC)

@Dragons flight, I've read your posting twice through looking for the bit of it which addresses the original question, and I haven't found it. --ColinFine (talk) 20:09, 26 September 2010 (UTC)

I think DF was trying to say that Baileys Irish cream is probably safe after 30 months (definitely after 2 years) if stored properly even in an opened bottle. This partly addresses one of the questions the OP asked. Df also quotes the website, which makes it clear the manufacturer guarantees the taste of Baileys Irish cream will remain for 2 years, even in an opened bottle, if stored properly. My interpretation of that Baileys is also saying that the 30 months is for an opened bottle stored under proper conditions as well although I'm less certain of this. This again partially addresses one of the OPs questions. Finally from DF's quotation it's clear that refrigeration is okay, which again partially addresses one of the OP's questions (or more accurately corrects one of their misconceptions). It isn't clear whether refrigeration is better, although it's clear the manufacturer suggests it's better then allowing it to get above 25 degrees C, which may or may not be relevent to the OP.

Actually from writing this I'm struggling to see how you can read the posting twice thorough and not see the quite useful info DF provided in response to the OP's questions and comments. It's true DF's answer wasn't perfect, and there are still parts of the OPs questions unanswered, but I don't think they said it was. And this is the RD, part of the process is that often each person may provide some info which may lead to the complete answer. Although often we don't even get a complete answer if there can really be said to be such a thing (particular since 'safe' is a fairly iffy word and will almost definitely depend on the precise conditions anyway so we can't really answer how long it will be safe for although info on precisely how it changes, e.g. at different temperatures obviously would be useful).

Nil Einne (talk) 21:50, 26 September 2010 (UTC)

Radioactive isotope of iron

When iron is irradiated with neutrons an isotope of iron is formed. This isotope is radioactive with a half-life of 45 days. A steel piston ring of mass 16 g was irradiated with neutrons until its activity due to the formation of this isotope was 10 µCi. Ten days after the irradiation the ring was installed in an engine and after 80 days' continuous use the crankcase oil was found to have a total activity of $1.85\times 10^{3}$ disintegrations per second. Determine the average mass of iron worn off the ring per day assuming that all the metal removed from the ring accumulated in the oil and that 1 Ci i equivalent to $3.7\times 10^{10}$ disintegrations per second.

I get 8.57 µCi for the activity after the ten days (i.e. immediately before it was installed in the engine). I have no idea how to do the rest. --MrMahn (talk) 07:04, 26 September 2010 (UTC)

Trying not to do too much Helping With Homework:

Basically, I think you're making this harder than it is; you don't need the activity after ten days, just after ninety (ten before installation + eighty while in there), which conveniently works out to exactly two half-lives. From there, you can work out what fraction of that activity is observed from the oil; that must also be the fraction of the original 16g that's worn off, and working out the per-day rate becomes trivial. --81.153.109.200 (talk) 07:29, 26 September 2010 (UTC)

Following your approach, I get 0.004 grams per day. Is this correct?--MrMahn (talk) 10:47, 26 September 2010 (UTC)

It's what I get (actually, the first time I tried I got 4 ng/day because I forgot about the micro in the μCi, which is the sort of thing I've been whining at my new A-level students for all month so *ahem*), and it's a nice round number so it seems like it should be right, but I wouldn't like to guarantee it.--81.153.109.200 (talk) 19:42, 27 September 2010 (UTC)

Could ancient Egyptians see or not see the three "aligned" volcano peaks of Mars?

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• The three aligned volcano peaks of Mars, where they visible to the ancient Egyptians who build three aligned major pyramids, or was that in complete coincidence?? 24.78.166.69 (talk) 11:40, 26 September 2010 (UTC)

That is even less probable the the Orion alignment. Orion Correlation Theory--Aspro (talk) 12:13, 26 September 2010 (UTC)

It's coincidence - unless the ancient Egyptians had discovered the telescope or were visited by Martian space travellers ! Martian volcanoes are certainly not visible to the naked eye, and even early telescopes only showed large, high contrast surface features such as the polar caps and Syrtis Major. Look at Huygens' 17th century drawing of Mars in our article on history of Mars observation to see what I mean. Gandalf61 (talk) 12:22, 26 September 2010 (UTC)

Prior to visits by 20th-century space probes, the site of Olympus Mons had been named by 19th-century terrestrial astronomers (using telescopes, of course) as "Nix Olympicus" (i.e. "Snows of [Mount] Olympus"), implying that they had spotted the white clouds that sometimes form adjacent to that mountain, but pre-probe no-one is known to have glimpsed the three (smaller) aligned volcanoes of the Tharsis Bulge, and I suspect that (terrestrial) atmospheric distortions would make that impossible with anything but the very latest and largest current telescopes. 87.81.230.195 (talk) 15:26, 26 September 2010 (UTC)

If they were replicating the Martian pattern, surely the largest pyramid would be off to the side from the line of three, corresponding to the relationship of Olympus Mons to the Tharsis volcanoes. Looie496 (talk) 16:44, 26 September 2010 (UTC)

Timber recovered from an archaeological excavation

A piece of timber has been recovered from an archaeological excavation and it is required to find its approximate age by measuring the radioactivity of the carbon-14 contained therein. For this purpose it may be assumed that the proportion of carbon-14 in the natural carbon of living wood is everywhere and at all times the same and that it begins to decay at death. If the number of disintegrations observed from 5 g of carbon prepared from the specimen is 21 per minute, how old is the specimen? (The proportion of carbon-14 to natural carbon in living wood is 1.25 in $10^{12}$ and the half-value period of carbon-14 is 5600 years. The mass number of natural carbon is 12.)

The mass of carbon-14 in the 5g of timber before it died was $5*{\frac {1.25}{10^{12}}}=6.25*10^{-12}g$
This equates to ${\frac {6.25*10^{-12}}{12}}=5.21*10^{-13}mol$
so $N_{0}=5.21*10^{-13}*6.02*10^{23}=3.14*10^{11}$
so $A_{0}=N_{0}\lambda =3.14*10^{11}*{\frac {\ln 2}{5600*60*24*365.2425}}=73.79$ disintegrations per second.
${\frac {A}{A_{0}}}=e^{-\lambda t}$
so ${\frac {21}{73.79}}=e^{-{\frac {\ln 2}{5600*60*24*365.2425}}t}$
solving this for t gives me a value which is clearly not correct.--MrMahn (talk) 11:59, 26 September 2010 (UTC)

First, whoever gave that question should be reprimanded for suggesting that "natural carbon" has a mass number of 12, or for using "natural carbon" to mean ¹²C. Secondly, you should use 14 (the molar mass of ¹⁴C) in the denominator of your second equation. And finally, if you get 73.79 (-~15% due to the error with 12/14 above), then 21/73 of the original material remain, or about 30%. In other words, a bit less than two half-times have elapsed, and the sample is ~10000 years old. --Stephan Schulz (talk) 12:16, 26 September 2010 (UTC)

Dendrochronology is far more accurate than carbon dating when intact timber is found.--92.251.191.21 (talk) —Preceding undated comment added 16:06, 26 September 2010 (UTC).

Yes, under ideal conditions. But dendrochronology is useless if only small pieces of wood are recovered (you need enough contiguous rings to match the known record), and less reliable in areas without good tree-ring records. TenOfAllTrades(talk) 17:02, 26 September 2010 (UTC)

Distance that aliens could detect earth with our level of technology

Given that Aliens exist with the same level of technology as ourselves, from what distance could they detect our solar system, (by Radial or Transit methods). And more specifically, it is likely that Jupiter, and then Saturn would be detected, at what distance could they detect Earth? —Preceding unsigned comment added by Csmiller (talk • contribs) 16:03, 26 September 2010 (UTC)

Not directly answering your question, but you may find this interesting. Interestingly it's Neptune that would be the easiest to detect, because of its influence on the dust in the Kuiper Belt. Also, our current level of technology doesn't allow for us to detect planets the size of those in our system, even Jupiter, at interstellar distances. Rojomoke (talk) 16:18, 26 September 2010 (UTC)

Aye, I saw that going past on my /. RSS newsticker. It prompted me to ask; I've been meaning to do so for a while. CS Miller (talk) 16:30, 26 September 2010 (UTC)

As our extrasolar planet article explains, current technology does in fact allow detection of planets much smaller than Jupiter. Capabilities have improved a lot over the past few years. Looie496 (talk) 19:47, 26 September 2010 (UTC)

Wouldn't any alien civilization have an easier time detecting all the radio frequency radiation that we're emitting than actually detecting the planets themselves? I thought that was the point of SETI (except in reverse; we're trying to detect aliens' radio signals). If they could pick up a couple episodes of Bill Nye the Science Guy they'd know far more about our solar system than they ever could by studying it at interstellar distances. Buddy431 (talk) 23:34, 26 September 2010 (UTC)

The Fermi paradox article has:

"Using our own experience as an example, we could set the date of radio-visibility for Earth as December 12, 1901, when Guglielmo Marconi sent radio signals from Cornwall, England, to Newfoundland, Canada. Visibility is now ending, or at least becoming orders of magnitude more difficult, as analog TV is being phased out. And so, if our experience is typical, a civilization remains radio-visible for approximately a hundred years. So a civilization may have been very visible from 1325 to 1483, but we were just not listening at that time. This is essentially the solution, "Everyone is listening, no one is sending."

WikiDao ☯ (talk) 23:45, 26 September 2010 (UTC)

Did anyone but the wanna be tyrannical business mogul Marconi claim to have heard that Morse code "S" consisting of three dots? How likely is it that he lied or was deceived by static in the earphone and his knowledge and expectation of what was to be sent? There was no proper protocol, which would have consisted of some trusted person handing the sender a previously unknown message to be sent. Edison (talk) 03:00, 27 September 2010 (UTC)

Guglielmo Marconi#Transatlantic transmissions has some discussion. It appears while there is some doubt about his earlier experiment, his later experiment is more well accepted. So the date is at most out by a few months Nil Einne (talk) 13:20, 27 September 2010 (UTC)

Re-asking Csmiller's question in terms of Buddy431's comment, how close would an alien civilization have to be to us in order to detect our radio emissions if their SETI equipment had the same sensitivity as ours? I recall reading that our strongest military radars might generate signals that would be just barely detectable at the nearest star systems. I'll try to find a reference, as I think that such information should be in the article. -- ToE^T 15:11, 27 September 2010 (UTC)

OK, here's a start. From David Brin's 2006 anti active SETI essay, SHOUTING AT THE COSMOS ...Or How SETI has Taken a Worrisome Turn Into Dangerous Territory:

... in a paper written by Dr. Shostak ..., even military radars and television signals appear to dissipate below interstellar noise levels within just a few light years.

-- ToE^T 15:54, 27 September 2010 (UTC)

http://www.space.com/searchforlife/seti_shostak_aliens_031023.html There are enough such radars that, at any given time, they cover a percent of the sky or so. The signal from the most powerful of these could be found at 50 light-years distance in a few minutes time with a receiving antenna 1,000 feet in diameter. Indeed, these military radars are the only signals routinely transmitted from Earth that are intense enough to be detectable at interstellar distances with setups equivalent to our own SETI experiments. Hcobb (talk) 16:07, 27 September 2010 (UTC)

is this star fruit? what is it?

Mystery tree

—Preceding unsigned comment added by Hemanetwork (talk • contribs) 16:04, 26 September 2010 (UTC)

Mostly definitely not starfruit, the fruit don't look at all like it (they don't even look like stars). Could be rambutan or lychee but the picture is fairly blurry Nil Einne (talk) 16:28, 26 September 2010 (UTC)

It's an arbutus, sometimes known as a strawberry tree. The fruits are edible once they reach the orange stage, by the way, although not very tasty. Looie496 (talk) 16:32, 26 September 2010 (UTC)

Looks like Arbutus to me. Can be wrong though, too blurry. Oops Looie's comment didn't show up until I've submitted mine. Sorry Looie.--Dr Dima (talk) 16:33, 26 September 2010 (UTC)

If you look up carambola, you will see the real star fruit. When the fruit is sliced across its width, the pieces look like star fish. Myles325a (talk) 02:10, 29 September 2010 (UTC)

motion of rocket.

A rocket is in outer space far from any planet when the engine is turned on.In the first second of firing the rocket eject 1/120 of its mass with relative speed of 2400 meters per second.(1).Calculate the rockets initial acceleration. (2). Suppose 3/4 of the initial mass,M of the rocket is fuel so that the final mass is M/4 and that the fuel is completely consumed at a constant rate in a total time of 90sec. (i)If the rocket starts from rest in our coordinate system,find its speed at the end of 90sec. (ii)what is the velocity of the last portion of ejected fuel relative to our reference system. —Preceding unsigned comment added by 41.218.236.95 (talk) 16:08, 26 September 2010 (UTC)

Please do your own homework.

Welcome to Wikipedia. Your question appears to be a homework question. I apologize if this is a misinterpretation, but it is our aim here not to do people's homework for them, but to merely aid them in doing it themselves. Letting someone else do your homework does not help you learn nearly as much as doing it yourself. Please attempt to solve the problem or answer the question yourself first. If you need help with a specific part of your homework, feel free to tell us where you are stuck and ask for help. If you need help grasping the concept of a problem, by all means let us know. DMacks (talk) 16:33, 26 September 2010 (UTC)

While I assume that this is a homework question, I can't imagine anyone posting one here right away unless they tried to solve it many times already.--92.251.191.21 (talk) 17:19, 26 September 2010 (UTC)

What DMacks means is, 'Please show us what you've tried so far, or how far along you've gotten'. It's Sunday night, the homework's due in the morning, what has he tried and where is he stuck? We'll help — but we want to know if it's a problem with understanding the concepts, difficulty interpreting a formula, or just a calculation error. TenOfAllTrades(talk) 17:35, 26 September 2010 (UTC)

Speaking in the general sense, I suspect you'll be wrong. In fact, I can't recall if there was ever any evidence '"physics magazine" guy' attempted his homework questions so we may have a whole lot of examples right there. Nil Einne (talk) 19:35, 26 September 2010 (UTC)

Perhaps the OP would care to read our article Tsiolkovsky rocket equation. -- ToE^T 14:21, 27 September 2010 (UTC)

Gold is the most electronegative of all the transition metals -- why isn't it a valuable organic catalyst like Pt or Pd or Rh or Os is?

Is it that it's too electronegative that it can't form metastable catalytic compounds like Pt can? John Riemann Soong (talk) 16:31, 26 September 2010 (UTC)

This type of catalysis is called "surface catalysis" or "Heterogeneous catalysis". It could be that while the electronegativity of the gold is favorable, the properties of the surface of the gold differ from other metals used. Also, consider the electron configuration of the elements in question, which will effect how they form coordinate bonds. Platinum, Palladium, and Nickel (Raney nickel) are all very common catalyst for these types of reactions, and look at the fact that the ALL have the same valence electron configuration, which means they probably tend to form the same bonds at the same angles and all of that. Consider that gold is in a group with copper and silver, NEITHER of which is terribly useful for this sort of catalysis. I am sure that electronegativity is a less important property here than coordination chemistry, which is dependant on the electronic environment around the atom. --Jayron32 05:06, 27 September 2010 (UTC)

Apparently, there are a number of recent papers on the use of gold as a catalyst [2]. The consensus seems to be that gold was not used simply because there was no research on it; organic chemists had long considered gold too inert and expensive. Someguy1221 (talk) 05:13, 27 September 2010 (UTC)

Reactions on surfaces deals with catalysis of this type; following blue links will give you some background on specific classes of surface reactions. The cost issue for gold hasn't been a concern for like 50 years, see the work of Vladimir Haensel on ultra-fine depositions of platinum on alumina surfaces (less than one part per thousand platinum); specifically used to keep cost down. There's no reason gold could not have been used in this manner. --Jayron32 05:24, 27 September 2010 (UTC)

Yellow copper compound

I react cuprous chloride (I'll tell you how I made it later) with sodium carbonate to get a yellow precipitate, a colorless solution, and fizzing. What is that yellow solid?

The cuprous chloride is made by reacting acidified cupric chloride with ascorbic acid.

The yellow solid dissolves in ammonia to make a yellow solution. It reacts with ascorbic acid to make cuprous chloride again. It reacts with hydrochloric acid to make some fizzing (excess sodium carbonate?) and a green cupric solution. It is not oxidized by sodium hypochlorite or hydrogen peroxide.

Thanks. --Chemicalinterest (talk) 20:09, 26 September 2010 (UTC)

You should try something called recrystallization, although you have to be careful (not to lose your product). Sometimes white powders for example have some discolouration cuz they have impurities. (Unless you do it already?) Although I guess if it really colours the ammonia solution yellow, it must be naturally yellow. John Riemann Soong (talk) 22:18, 26 September 2010 (UTC)

Cuprous chloride is colorless and insoluble. The yellow powder is insoluble as well. Recrystallation would not seem to work well. Also, why does it turn back to CuCl when reacted with ascorbic acid? --Chemicalinterest (talk) 11:11, 27 September 2010 (UTC)

Well you can dissolve it in ammonia or HCl and neutralise it with the other. But you don't need to do that. Ascorbic acid can be a pro-oxidant as well as an antioxidant under certain circumstances. I guess your results exclude metallic copper. Hmm. What about copper(I) oxide? It can be yellow if your particles are small enough and ascorbic acid reactions tend to produce colloidal nanoparticles. John Riemann Soong (talk) 15:24, 27 September 2010 (UTC)

Copper(I) oxide in ammonia is colourless, but I wonder if your precipitate is in fact, two different products or just one. You could also set up an electrochemical cell with a diagnostic half-cell (say metallic copper) and measure the voltage -- it won't be standard state but you could see in which way a redox would flow. John Riemann Soong (talk) 15:25, 27 September 2010 (UTC)

is CuCl soluble in neutral salty water?

After all, it's the Cl- and not the H+ that makes it soluble right?

Also why couldn't aqua regia also be made with NaCl added? The Cl- helps complex the oxidised gold. John Riemann Soong (talk) 22:26, 26 September 2010 (UTC)

Both questions can be quickly answered by considering the solubility of sodium chloride in water, and comparing it to the solubility of hydrogen chloride in water. Physchim62 (talk) 00:53, 27 September 2010 (UTC)

Sodium chloride is much less reactive. Does it react with manganese dioxide to produce chlorine? --Chemicalinterest (talk) 11:09, 27 September 2010 (UTC)

Vehicle damage from hitting a pedestrian?

I've been trying to find some kind of data or estimate regarding the average dollar amount for damages to vehicles in pedestrian accidents, but can't find anything. Can anyone give me any info on this or point me where I might be able to find this type of information?

What I'm ultimately trying to figure out is if $200 is a lot of damage for a vehicle after hitting a six year old child, or how fast a vehicle might have been travelling to have caused $200 damage in such an "accident"? Also, can the dollar amount written on the police incident report be considered an accurate assessment of such damage, or do they tend to be way off (would I be better off trying to get 25 y/o court documents for this info, or can I be reasonably sure the on-scene report I have is fairly accurate)?

(Is this a science question?)

Thanks in advance for any ideas. bcatt (talk) 22:35, 26 September 2010 (UTC)

I don't know about 25 years ago, but nowadays you'll be lucky if $200 will pay for a windshield replacement. A smashed headlight might be similarly expensive. If the airbags deploy (I'm not sure how likely this is.) you could be looking at a good deal more than $200 to replace them. APL (talk) 00:23, 27 September 2010 (UTC)

Yeah i was going to say the same. My dad hit a dog a few years ago doing about 50km/h and it did $1200 damage to a petty crappy old car and it was just the headlight, bumper and a bit of panel damage. As for the rest, it would depend on a lot of factors and I think it would border on "legal advice". Vespine (talk) 00:31, 27 September 2010 (UTC)

Agree with Vespine in that there are just too many factors to consider in this question. I don't think an "average dollar amount for damages to vehicles in pedestrian accidents" is a meaningful average. The amount will range from zero to the assessed value of the vehicle (i.e., write off), and those of us who are drivers will know that the assessed value of vehicles in itself ranges over several orders of magnitude. I don't think the OP is looking for legal advice, as I don't think any court would take an "average damage" into consideration: I interpreted this as a request for input into some sort of economic debate. Physchim62 (talk) 00:49, 27 September 2010 (UTC)

Just re-painting the front bumper – not replacing, just painting – can cost a couple of hundred dollars and up. (This post in a BMW owners forum noted a cost of $800 for a BMW-certified collision center to respray a front bumper in 2009.) This discussion among Ford owners notes specifies three hundred to almost a thousand dollars to repaint the hood of a Ford Explorer, in 2002. In other words, just modest paint damage can cost a few hundred dollars to repair; prices go up if there is damage across more than one surface panel. Replacing any one of a damaged bumper, grille, or headlamp assembly can again run into the low hundreds these days (thousands, if it is a specialty imported vehicle). It's even possible that $200 is given as a sort of nominal 'minimal damage' amount — it's virtually impossible to get out of a body shop without spending at least that much, even for very minor dents and scrapes. Overall, the dollar cost of repairs is a very sloppy measure of how much 'damage' was done in an accident, and it has an even less direct relationship with vehicle speed or the other parameters of the accident itself. TenOfAllTrades(talk) 02:28, 27 September 2010 (UTC)

I don't know if it is the same in the USA, but a few years ago my wife scraped a car door and we were quoted £800 for it to be repainted. The car was fairly old so we decided jsut to paint it ourselves. The guy then said "I didn't realise that it's non insurance, I'll do it for £200". There is a huge markup on insurance work in the UK. -- Q Chris (talk) 12:06, 27 September 2010 (UTC)

Yep. Similar story in Australia. Want a cheap job? Tell the panel beater it a non-insurance job and that you will be paying cash. HiLo48 (talk) 12:40, 27 September 2010 (UTC)

Thanks to everyone who responded. It's true I'm not looking for legal advice, though it's not part of an economic debate either. I have this on-site incident report from when I was hit by a car in a crosswalk about 25 years ago; it states the vehicle damage as $200, and it just got me curious what exactly that means. I also have no memory of actually being hit - just stepping into the crosswalk then lying on the ground in shock - so I was also curious if I might be able to use that amount to guess at what speed the driver may have been going...part of sort of a self-therapy type thing, I guess (probably better to get the court records if I want to delve into it anyway). I think TenOfAllTrades is probably right about $200 being a "minimal damage" amount. Anyway, thanks again to all. 24.69.99.45 (talk) 19:59, 27 September 2010 (UTC)

According to [3] prices have doubled between August 1985 and August 2010. (Assuming that you are using US dollars) CS Miller (talk) 10:37, 28 September 2010 (UTC)