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November 20

What's the solution to Haldane's dilemma?

I read the article and still don't understand. Sagittarian Milky Way (talk) 03:23, 20 November 2012 (UTC)[reply]

The article doesn't seem to provide a great or simple overview of the topic, but from what I understand, the "dilemma" is due to some people's interpretation that a mathematical model can be used to make a vague and nonquantitative prediction in contradiction to observations. It seems to me that the "resolution" is not to trust the flawed mathematical formulation, because it makes vaguely incorrect predictions. Where are you getting stuck? Lots of scientists have designed models, or created mathematical equations, that turned out to be wrong. Even more scientists have produced proper, accurate models, that are misunderstood and misused by others to make erroneous claims. Perhaps you'd do better to ignore the "dilemma" and simply read our better article about population genetics.Nimur (talk) 05:23, 20 November 2012 (UTC)[reply]

I'll read that, tomorrow. Interesting I'm sure. Sagittarian Milky Way (talk) 07:05, 20 November 2012 (UTC)[reply]

Yup: from my understanding of the article, the 'solution' is that the dilemma never existed in the first place: Haldane was making assumptions that don't accord with real genetic/evolutionary processes, so his conclusions were invalid. AndyTheGrump (talk) 05:31, 20 November 2012 (UTC)[reply]

See [1]. The number of mutational changes that occur really is much greater than the number that become fixed - most of the time, random mutations are neutral mutations. There is also the truncation selection, which I would say is relevant in so-called "junk DNA" - a concept which is now growing old, but which at least accurately reflects the point that regulatory sequences are prone to constant fine-tuning and there are a bazillion different ways to get the same effect. (You can have a 6 base pair motif that isn't exact anywhere in a fairly loose region of DNA and the outcome may work out to be the same pattern of regulation) So it doesn't really matter how evolution serves you up a "strong COMT promoter" or a "weak COMT promoter", depending on whether warriors or worriers are doing better against the hyenas this season. There's no need for fixation of one precise change because there are so many ways to strengthen or weaken the binding of a particular protein to the promoter. Wnt (talk) 05:35, 20 November 2012 (UTC)[reply]

Well then the article should say that then. It's actually very simple. You're left with the impression that a resolution hasn't been found in 55 years despite extremely embiggened knowledge of genetics and the fact that scientists like explaining things. Besides "we know which side is correct" (the fossil record/radioisotopic dating/genetics and all that) (which is not mentioned by the way).

Hey, are there any other paradoxes which we know/knew which side is correct but don't know why? Sagittarian Milky Way (talk) 07:01, 20 November 2012 (UTC)[reply]

The article clearly needs work but it seems to accurately report that the dilemma is no real dilemma. For example right in the LEDE it says:

Contrary to creationist claims, Haldane's dilemma is of no importance in the evolutionary genetics literature

....

Haldane stated at the time of publication "I am quite aware that my conclusions will probably need drastic revision", and subsequent corrected calculations found that the cost disappears. He had made an invalid simplifying assumption which negated his assumption of constant population size, and had also incorrectly assumed that two mutations would take twice as long to reach fixation as one, while sexual recombination means that two can be selected simultaneously so that both reach fixation more quickly. The creationist claim is based on further errors and invalid assumptions

Nil Einne (talk) 14:35, 20 November 2012 (UTC)[reply]

As far as I can see just having had a quick look there is a basic flaw in that in reality the main competitors for an individual in a species are other members of that species. They do not normally compete with nature, they compete with each other. Thus his rate of death does not matter much normally when the environment is only changing slowly. Of course all bets are off nowadays that humans are around and killing everything in sight but catastrophes like that aren't the norm. Dmcq (talk) 10:12, 20 November 2012 (UTC)[reply]

And when rapid change occurs, populations do become extinct. Antibiotics wouldn't be much use if that wasn't the case. Millions of populations of bacteria are destroyed before one becomes resistant. In the section Origin of the term "Haldane's Dilemma" it says if an environmental change occurs that necessitates the rather rapid replacement of several genes if a population is to survive, the population becomes extinct. That's not a dilemma, that's reality. Ssscienccce (talk) 01:39, 21 November 2012 (UTC)[reply]

Bear pig

What animal is equally related to a bear and a pig?GeeBIGS (talk) 05:47, 20 November 2012 (UTC)[reply]

What do you mean "Equally related"? I'm not sure I follow your question... --Jayron32 05:53, 20 November 2012 (UTC)[reply]

Did they have a common ancestor? What did it look like? Similar to anything extant today?GeeBIGS (talk) 05:56, 20 November 2012 (UTC)[reply]

Ah. It would have to be a pretty old ancestor. Taxonomically speaking, the smallest grouping that both bears and pigs share is the clade Laurasiatheria. I'm not entirely sure what the nearest common ancestor for that clade is, but it's got to be a long time ago, among mammals. --Jayron32 06:08, 20 November 2012 (UTC)[reply]

What would a hybrid possibly look like?GeeBIGS (talk) 06:00, 20 November 2012 (UTC)[reply]

I don't know that bears and pigs are genetically close enough to hybridize. Most hybrids need to share a genus, usually. Bears and pigs don't even occupy the same order (Carnivora vs. Artiodactyla). --Jayron32 06:08, 20 November 2012 (UTC)[reply]

You can answer many questions of this type pretty easily on Wikipedia (I haven't looked for any recent updates on NCBI). Look up bear. Look up pig. Look up class, order, etc in the table at right. Both are mammals; the bear is order Carnivora, the pig is order Artiodactyla. Now look those up and you'll see both of these are in superorder Laurasiatheria. Now - look for anything in Laurasiathera that is not in Carnivora or Artiodactyla (well, Cetartiodactyla and Ferae, once you look at the tree) - that gets you Perissodactyla, any odd-toed ungulate, as a potential answer, though as the branch is unresolved in the taxonomy shown it's still possible it's closer to one or the other. Oh, and yes -- the way you phrase the question, anything more distantly related also qualifies, plants, yeasts, sea urchins etc. But looking for the closest animal equally related to both is the more interesting question. :) Wnt (talk) 06:09, 20 November 2012 (UTC)[reply]

I am satisfied with a tapir for now.GeeBIGS (talk) 06:20, 20 November 2012 (UTC)[reply]

It's an entirely invalid assumption to just pick an animal from a separate order (tapir) but within the same superorder/class and assume that it's related with equal distance from both starting point. And furthermore, the OP jumps from asking what the actual equally related animal is to what would a possible hybrid look like -- these may very well be two totally different things, because closest same relateness might have very little to do with outward appearance (phenotype). DRosenbach ^{(Talk | Contribs)} 13:55, 21 November 2012 (UTC)[reply]

Depending on who asked the question, the answer they were looking for may have been Al Gore. But taxonomically, Jayron is right. However the earliest Laurasiathere certainly did not look like a Tapir, and probably looked more like a rat or an opossum. μηδείς (talk) 17:28, 20 November 2012 (UTC)[reply]

GeeBIGS—you ask the question, "What would a hybrid possibly look like?" Is the question what actual animal a possible hybrid would look like? Or is the question what nonexistent animal a "hybrid" would look like? I believe there is software that makes an image metamorphose into another image. It does so over many steps. One of the middle steps would be a credible transitional animal between bear and pig. Bus stop (talk) 17:39, 20 November 2012 (UTC)[reply]

An attempt at this research was reported 2008-4-1.[2] Of course, this is discredited; it would be unethical to use cells from a bear.[3] Wnt (talk) 18:18, 20 November 2012 (UTC)[reply]

Earlier, unconfirmed reports had existed for almost two years at that point. --Jayron32 19:32, 20 November 2012 (UTC)[reply]

Kidding aside, I'm loath to rule out the possibility of such a hybrid, even at the genetic level. Our article on polyploidy introduces me to the curious case of the Plains viscacha rat, which has been proposed as a recent polyploid. Research into the animal might reveal mechanisms by which dosage compensation for the duplicated genome could have succeeded. If you have a means for polyploids to survive, then it becomes feasible for bear-pig fusion nuclei to survive, except for some relatively narrow mechanisms of reproductive isolation which conceivably could be worked out in cell culture of some embryonic stem cell precursors. In the long term bottom line, there is no theoretical reason why bears and pigs can't be cleanly hybridized; it is strictly a technical problem (though a substantial one) and when push comes to shove you could build a synthetic combination up gene by gene if you had to. By comparison, there was no logical reason why somatic cell cloning ever had to be possible; that was dumb luck/excessive simplicity in evolution. Wnt (talk) 22:40, 20 November 2012 (UTC)[reply]

A bear is about as closely related to a pig as a person is to a rabbit. Good luck with that. μηδείς (talk) 23:53, 20 November 2012 (UTC)[reply]

Someone once crossed an owl with a goat, don'cha know. ←Baseball Bugs ^{What's up, Doc?} carrots→ 06:36, 21 November 2012 (UTC)[reply]

Any such attempt would be certain to make both the owl and the goat rather cross, although I have seen the occasional old goat with owl eyebrows: [4]. :-) StuRat (talk) 07:23, 21 November 2012 (UTC) [reply]

Hmmm... evidently you've never heard of the famous hootin' nanny. ←Baseball Bugs ^{What's up, Doc?} carrots→ 21:00, 21 November 2012 (UTC)[reply]

How can I separate mixture of water and ammonia?

Not sure if boiling or freezing works--124.172.170.234 (talk) 06:39, 20 November 2012 (UTC)[reply]

Probably neither directly; ammonia as a gas is hard to seperate out by distillation. Instead, acidify the solution careful to the equivalence point, with (for example) hydrochloric acid. This should allow you to distill the water off from the resultant ammonium chloride. --Jayron32 06:50, 20 November 2012 (UTC)[reply]

If it's a reasonably concentrated solution, there's a pretty high vapor pressure of ammonia over the solution especially when warmed (see ammonium hydroxide, and in general you can boil some ammonia out (see ammonia#Properties. But the answer really depends really what the questioner's goal is (actually "separate the two components" vs "get some pure water from it" vs "get some pure ammonia from it"). Jayron's got the right start for the old industrial process (see ammonia#Synthesis and production). DMacks (talk) 08:36, 20 November 2012 (UTC)[reply]

Apparently ammonia and water don't form an azeotrope [5] - it is theoretically possible to do the distillation. But the relative ease with which ammonia can be converted to solid with HCl and back to gas with CaO means that it's not worth the trouble. Wnt (talk) 22:48, 20 November 2012 (UTC)[reply]

9th period

Are there any relativistic predictions for the 9th period of the PT (Z > 172), just as there are for the 8th (Pyykkö's)? (Note that Z = 173 is not actually predicted to be an end of the PT anymore – see Periodic table.)

Double sharp (talk) 07:16, 20 November 2012 (UTC)[reply]

Is the table you linked the Pyykkö model?--Jasper Deng (talk) 07:23, 20 November 2012 (UTC)[reply]
- Yes. Double sharp (talk) 07:46, 20 November 2012 (UTC)[reply]
  - Well, that model seems to make some predictions about some otherwise-period-8 elements being in the 9th period as in this table. I don't know of any predictions for other elements in the 9th period.--Jasper Deng (talk) 23:47, 20 November 2012 (UTC)[reply]
    - I wouldn't really call it period 9 – they're filling 9s and 9p_1/2 only because these are relativistically stabilized and have energies closer to that of 8p_3/2. Are there any predictions past Z > 172? Double sharp (talk) 05:13, 21 November 2012 (UTC)[reply]
      - Actually, that probably would be better called the 10th period, per Fricke's paper (see WT:ELEM for all the discussion). Double sharp (talk) 15:07, 24 November 2012 (UTC)[reply]

How much more diverse would the species be if we hadn't gone through the Toba catastrophe?

Some alleles were lost forever, right? But I mean, how much more freaking diverse do you want the human race to be? Would we start having elf ears or white hair at youth or tribes with very strong bonds without ritual circumscision that only like making love? Would the rarer traits like Zulu veins or Sarah Baartman's anterior be more common? Perhaps some more blood types or major histocompatibility complexes or body odors? And about how close are we to filling the a full subspecies of genetic diversity? Not much, maybe? How much of the Homo sapiens genetic field is extinct? Sagittarian Milky Way (talk) 07:44, 20 November 2012 (UTC)[reply]

I am not sure which I find more troubling, your fear of elves or of people who like sex. μηδείς (talk) 17:17, 20 November 2012 (UTC)[reply]

Is this a joke? Elf ears make anyone less sexually attractive, that doesn't mean I'm afraid of them. And the world would be better with fewer pick up artists and wars of aggressionSagittarian Milky Way (talk) 07:41, 22 November 2012 (UTC)[reply]

(ec) As a first point, the idea that the Toba supereruption caused a genetic bottleneck doesn't get much support nowadays -- there was very likely a major bottleneck, but much earlier than that. In any case, genetic diversity does not necessarily show up as overt physical differences. For example, pygmies in West Africa show a considerably higher level of genetic diversity than a group of randomly chosen Europeans. Generally, though, if you want to see the effects of diversity, look at sub-Saharan Africa -- that's where diversity is greatest. Looie496 (talk) 17:19, 20 November 2012 (UTC)[reply]

The article on Toba doesn't seem as deinitive as you are. Rmhermen (talk) 19:11, 20 November 2012 (UTC)[reply]

I didn't think I was being all that definitive, but, well, the article hasn't really kept up with events. The Toba eruption is pretty clearly dated to about 70000 years ago. It has recently become clear that the rate of human genetic change is about half as high as the value that was used for a long time (PMID 22965354), and when you revise dates accordingly, the numbers don't even come close to working out. Looie496 (talk) 19:32, 20 November 2012 (UTC)[reply]

Stop talking about my anterior please. Sarah Baartman (talk) 18:02, 20 November 2012 (UTC)[reply]

(!) μηδείς (talk) 23:45, 20 November 2012 (UTC)[reply]

Someone's confused about what anterior means. ←Baseball Bugs ^{What's up, Doc?} carrots→ 20:58, 21 November 2012 (UTC)[reply]

[6] Sarah Baartman (talk) 21:21, 21 November 2012 (UTC)[reply]

Sure that's a big butt, but if you see how out-of-proportion butts can get in New York you'd save your shock for the homo sapienses with the vulvas of aliens!. Sagittarian Milky Way (talk) 07:41, 22 November 2012 (UTC)[reply]

What are "Zulu veins"? Bielle (talk) 23:23, 23 November 2012 (UTC)[reply]

free electrons per atom

How do you calculate ab initio the number of free electrons per atom. For example, Tin is found in both the +2 and +4 oxidation states. Which ones do you use for a resistivity or a mean free path calculation? However, for graphite, based on what I know about organic chemistry, it would be 6 pi electrons per six carbons, or 1. (Yet carbon is frequently found in a +4 oxidation state, like carbon dioxide). Which oxidation states do you use in the context of electrons being on the scale of the fermi velocity? 71.207.151.227 (talk) 11:04, 20 November 2012 (UTC)[reply]

Help! This is urgent. Why is it so hard to find such a list of such simple values on the internet? 71.207.151.227 (talk) 11:12, 20 November 2012 (UTC)[reply]

A better question might be, why do you think oxidation numbers have anything to do with the number of free electrons in a solid? The Fermi energy, by itself, tells you the effective density of free electrons. See for example this discussion [7]. There is also a table of Fermi energies / velocities [8]. Dragons flight (talk) 11:27, 20 November 2012 (UTC)[reply]

I'm supposed to calculate the Fermi velocities ab initio, so that won't do. 71.207.151.227 (talk) 11:57, 20 November 2012 (UTC)[reply]

You can calculate Fermi energy / free electron density from the resistivity, if you know that. What are you allowed to use, cause oxidation states probably aren't going to be what you want. Dragons flight (talk) 12:06, 20 November 2012 (UTC)[reply]

heat capacity of a plasma

What's the model for the heat capacity of a plasma? Is C_P still C_V + R? Are there lattice vibrations in a plasma? 71.207.151.227 (talk) 11:55, 20 November 2012 (UTC)[reply]

A plasma is generally approximated as a gas of partially or fully ionized nuclei superimposed with a gas of free electrons. There is no lattice in a plasma. Dragons flight (talk) 12:29, 20 November 2012 (UTC)[reply]

Isolated systems

Do true thermodynamically isolated systems occur in biology? If not, what about anywhere in the natural world? Are even closed systems possible in biology? I've heard an egg could be. — Preceding unsigned comment added by 94.116.147.229 (talk) 15:56, 20 November 2012 (UTC)[reply]

No, entirely isolated systems do not exist, they are a theoretical concept that can however be used as an approximation of a naturally occurring system in some cases. The only exception is the universe itself, because it by definition contains all matter and energy in existence, so that excludes the possibility of external interaction. And yes, an egg for all practical purposes can be considered a closed system, though I think that very small particles like hydrogen molecules could leak through the shell. - Lindert (talk) 16:13, 20 November 2012 (UTC)[reply]

Read Thermodynamic system#Isolated_system and Isolated system. Sarah Baartman (talk) 16:38, 20 November 2012 (UTC)[reply]

If an egg was a closed system, you wouldn't be able to hard-boil it! Remember, a closed system doesn't just prevent transfer of matter, but of heat/energy as well. 209.131.76.183 (talk) 18:37, 20 November 2012 (UTC)[reply]

No, that's an isolated system. A closed system can exchange energy but not matter. --Jayron32 19:30, 20 November 2012 (UTC)[reply]

Actually an egg is not a closed system at all, the chicks inside breathe through the shell. There are thousands of microscopic pores through which carbon dioxide and oxygen are transported. I couldn't find that information in our wiki article after a brief search. Vespine (talk) 01:02, 21 November 2012 (UTC)[reply]

Fuel consumption of engines

For small engines the fuel consumption is measured in g/kWh. If we multiple the SFC with the energy density of the fuel (44.4MJ/kg for gasoline in this case) we get the following expression: 300 g/kWh * 44400MJ/g = 13MJ/3.6MJ. Since the units cancel out we get a ratio, and it appears to be the inverse of the efficiency of the engine. My question is: is this actually the inverse of the efficiency? I double checked my work and everything seems correct but I'm having a hard time convincing myself that calculating the efficiency is as simple as a single multiplication.A8875 (talk) 16:29, 20 November 2012 (UTC)[reply]

Sounds pretty right, why would you expect it to be more complex? Efficiency is simply how much total energy is converted to usable energy. You have total energy on one side (the fuel) and usable energy on the other (the engine output). The hard part might have been calculating the g/kWh in the 1st place, but that's been done for you. Vespine (talk) 00:54, 21 November 2012 (UTC)[reply]

Your calculation is perfectly valid, and the answer you obtained is not unreasonable (equiv to ~ 27%). Note that when looking up the energy density of fuels it is customary to use the Low Heat Value and not the High Heat Value, as a gasoline engine does not condense the H₂O vapour in the exhaust. 44.4 MJ/kg is a reasonable value for gasoline LHV. Some people who have studied the otto air cycle formula in a physics course expect gasoline engines to have an efficiency around 50% as predicted by the otto formula. However, practical gasolines engines are nearer 25% due to a great many factors such as non-zero combustion time, heat loss to coolant, friction, etc. Keit 60.230.228.87 (talk) 00:57, 21 November 2012 (UTC)[reply]

Thanks, you two. I double checked my math a dozen times but still wasn't sure so I thought I'd get some expert opinion. I'm liable to making dumb mistakes, especially on fundamental concepts like this one.A8875 (talk) 01:11, 21 November 2012 (UTC)[reply]

There are a obvious error in your calculation, 300 g/kWh * 44400MJ/g should probably read 300 g/kWh * 44400 J/g. The result seems OK. Gr8xoz (talk) 16:27, 21 November 2012 (UTC)[reply]

Momentum transfers

I have absolutely no clue how to solve this type of problem. How do I figure out velocities after a collision or before it from the equation $m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$ ? This in in one dimension. --Melab±1 ☎ 16:44, 20 November 2012 (UTC)[reply]

Conservation of energy, or if it isn't conserved you need to specify an additional condition (e.g if the masses stick to each other after the collision that imples a relation between the velocities after the collision). Count Iblis (talk) 17:13, 20 November 2012 (UTC)[reply]

It is a totally elastic collision and the masses are not necessarily equal. I'm looking for the general solution. I tried solving for

v_{1}

but that did not work. --Melab±1 ☎ 17:18, 20 November 2012 (UTC)[reply]

m1 u1 + m2 u2 = m1 v1 + m2 v2

Totally inelastic collision: v1 = v2, so we have:

v1 = (m1 u1 + m2 u2)/(m1 + m2)

Count Iblis (talk) 17:25, 20 November 2012 (UTC)[reply]

But it can't be known for sure that

u_{1}=u_{2}

or that

m_{1}=m_{2}

. It could be that

u_{1}\neq u_{2}

or

m_{1}\neq m_{2}

or both. --Melab±1 ☎ 17:34, 20 November 2012 (UTC)[reply]

If the collision is totally inelastic, then the velocities after the collision are the same. Count Iblis (talk) 17:42, 20 November 2012 (UTC)[reply]

You can't solve for two variables if you only have one equation. This problem can't be solved without more information. Looie496 (talk) 17:44, 20 November 2012 (UTC)[reply]

Well, the problem I am working with only gives me the masses and their velocities prior to colliding. It is an elastic collision. They don't stick together. --Melab±1 ☎ 17:45, 20 November 2012 (UTC)[reply]

Have a look at Elastic_collision#One-dimensional_Newtonian. There are worked examples. You have TWO equations, one for conservation of momentum (which you wrote), and another for conservation of energy (which you did not). Using both, you can solve for new velocities, regardless of how the initial masses and velocities compare. The algebra can get rough, but our article also describes a nice trick of changing the frame of reference, so that one of the initial masses has zero velocity. SemanticMantis (talk) 17:56, 20 November 2012 (UTC) P.S. I have no Idea what Iblis is thinking above. Of course the velocities can be different after collision... [reply]

It did say inelastic at one point. I guess my edit got through after his. So does the

x_{1}

and

x_{2}

on that page equal

u_{1}t

and

u_{2}t

, respectively? Even still, I can't understand it. --Melab±1 ☎ 18:05, 20 November 2012 (UTC)[reply]

Apologies, I misread "elastic" for "inelastic". In that case, you have conservation of energy, but it is not recommended to use the equation for energy conservation, because the algebra then gets messy. Instead, you work in the center of mass frame where the total momentum is zero. Suppose we are in the center of mass frame, then the energy of two particles with momenta p1 and p2 with masses m1 and m2 is given by:

E = p1^2/m1 + p2^2/m2 = (p1^2) (1/m1 + 1/m2)

because in the center of mass p1 = -p2. Energy conservation thus implies that the magnitude of the momentum of the partices is the same before and after the collision.

The velocity of the center of mass is

u* = (m u1 + mu2 )/(m1+m2)

The initial velocities in the center of mass system are thus:

u1' = u1 - u*

u2' = u2 - u*

The two momenta in the center of mass frame are

p1 = m1 u1'

and

p2 = m2 u2'

they are equal in magnitude and they add to zero, so p1 = -p2

This means that after the collision, the momenta reverse sign, so particle 2 will get the momentum that particle 1 had and vice versa. So, we have:

m1 v1' = p2 = m2 u2' = m2 (u2-u*) ---->

v1' = m2 (u2-u*)/m1

The velocity of particle 1 after the collision in the original frame is thus:

v1 - u* = m2 (u2-u*)/m1 ------>

v1 = u* + m2 (u2-u*)/m1

Count Iblis (talk) 18:13, 20 November 2012 (UTC)[reply]

Or, if you want to avoid using the center of mass solution, you can use the energy conservation (not as messy as Count Iblis implied). Start with the energy conservation

m1 u1^2 + m2 u2^2 = m1 v1^2 + m2 v2^2

and rearrange the terms as

m1 (u1^2 - v1^2) = m2 (v2^2 - u2^2)

Open each parenthesis as a product of two factors

m1 (u1 - v1)(u1 + v1) = m2 (v2 - u2)(v2 + u2) (Equation 1)

use the momentum equation

m1 u1 + m2 u2 = m1 v1 + m2 v2

rearranged as

m1 (u1 - v1) = m2 (v2 - u2) (Equation 2)

to substitute the factors m1 (u1 - v1) into the equation 1 above and get

m2 (v2 - u2)(u1 + v1) = m2 (v2 - u2)(v2 + u2)

which simplifies to

u1 + v1 = u2 + v2 (Equation 3)

Equations 2 and 3 above now form a set of linear equations that can be easily solved.

Dauto (talk) 19:00, 20 November 2012 (UTC)[reply]

I have the feeling that it gets even easier if the center of mass is stationary in the frame of reference, as in m1 u1 + m2 u2 = m1 v1 + m2 v2 = 0. - ¡Ouch! (^{hurt me} / _{more pain}) 15:52, 22 November 2012 (UTC)[reply]

Recycling 10, 20, or 30 year old paper

Hello, I have some old documents in boxes, and these papers are probably 10, 20, or 30 years old. I intend to recycle all of these documents, and I was wondering, can they even reuse the fibers from paper that old? 2A02:AF8:1:3500:0:0:0:9868 (talk) 21:30, 20 November 2012 (UTC)[reply]

Yep. They have different grades of paper and know which fibers can be used for which grade. Trio The Punch (talk) 23:25, 20 November 2012 (UTC)[reply]

Area of land required to feed working horses on arable farm

I have seen somewhere a statistic indicating the proportion of the cropped area (ie excluding fallow) of a typical arable farm (worked by horse power) which was required to grow the crops required to feed the horses working the farm, the point of interest being that the introduction of tractors powered by fossil fuels led (in addition to general improvements in labour productivity) to release of that land for production of crops for sale off the farm. Does anybody know what that proportion was? (I am interested in comparing it with the proportion of bio-ethanol (or similar) produced from land worked by a tractor which would be required to power the tractor.)Peter MacLaren (talk) 21:41, 20 November 2012 (UTC)[reply]

This is a very interesting and very difficult question! There is certainly no general, one-size-fits-all answer, but I will be interested to hear other responses. You can google around a bit, and see estimates at that are several orders of magnitude apart. The land needed to sustain a horse depends on how hard it's worked, what crops is grown, what the soil type is and the biome. These will restrict the plants, growing seasons, and general agricultural biodiversity and cropping systems available. Are you interested only in the horse, or do the humans have to survive off the farm too? You might be interested in 40 acres and a mule, or Three acres and a cow (the implied agricultural estimates, not the politics). Your more general question is currently the subject of intense research, funded by many governments and corporations. For starters, check out Life cycle analysis.

This does not directly answer the question, but when Ford were selling the original Fordson farm tractor back in the 1920's, the sales kits provided to dealers by Ford compared the economics of the tractor against the use of horses for ploughing, reduced after explanation to a dollar for dollar comparison. This was somtimes published in Ford's advertising, so try searching in old farm magazines. Quite likely old farm magazines and manuals from the 1920's (when tractors were competing not so much against each other but with horses) have the information you want. National Libraies usually have them, some have converted them to PDF and put them on line. I have an old Fordson manual somewhere that has a summary of the comparison - if I remember where I put it, I'll post again. Keit 60.230.228.87 (talk) 01:08, 21 November 2012 (UTC)[reply]

Page 3 of my Fordson Model F (1916) tractor owner's manual has the following text under the heading Horse and Horseless Farming:-

The harness and whiffletrees for an eight-horse team cost more than a Fordson Tractor. Yet the eight won’t do more work.

The eight horses cost double the price of the Fordson, and that at the (current) low price of horses.

Grooming eight horses once a day at 15 minutes a horse takes two hours. Watering and feeding, another hour. Harnessing and Un-harnessing, hitching up and unhooking, leading from barn to implement, etc., take yet another hour. Four hours’ work has been lost without expenditure of any energy in productive work.

A Fordson can be filled with water, fuel and oil, and thoroughly gone over in half an hour.

A Fordson can be worked continuously day and night through all the seasons of plowing, seeding, haying, harvesting.

Horses cannot be humanely worked more than eight hours in the heavier operations or ten in the lighter.

Fordsons are not troubled with flies, heat or hard ground. Horses suffer terribly and die in appalling numbers when hard worked on hard land in hot weather.

A Fordson can do all that horses can do, as well as horses can do it and belt work besides.

It takes a few hours to make a Fordson.

It takes three years’ time and three years’ care (some horsemen say five years) to make a work horse. At any time in those three years the colt may die and be a total loss.

A Fordson eats only when it is engaged in productive work.

Horses eat 365 days a year.

A Fordson makes every acre of the farm a source of profit.

An eight-horse team withdraws 40 acres from the farm’s return to feed itself.

The following section titled "Plowing Acreage and Speed Data" goes on to explain how to calculate how much land you can plow per 10 hour day. For the Forson Model F, it works out to be 7.6 acres for typical conditions - the implication is that an 8-horse team would do less. On Page 7 it says that with a "special seeding machine", a Fordson can plant corn at 3 hours per acre and bale 30 tones of hay per day, and that if necessary in rush periods a Fordson can be worked (presumably with 2 or more drivers) day and night, whereas horses must be allowed time for feeding and sleeping etc. Thus, you can consider that under rush circumstances (eg getting work done before the rains come) a Fordson does the work of 24 horses, which require 120 acres for feed.

On Page 8 it says a Fordson all up (depreciation, fuel, oil, wages for driver, etc) cost a farmer $0.95 (this is 1916 data don't forget, based on tractor purchase price $880, horse $150, both have a working life of ~5000 hours) to plow an acre, whereas a horse team will cost $1.46 to plow an acre, and confirms that an 8 horse team will in 8 hours (not counting time to feed and harrnes) do the same plowing work as a Fordson in 10 hours (not counting the 0.5 hour daily fueling and checking).

Modern tractors could be expected to considerably outperform a 1916 Fordson, and last very considerably longer than 5000 hours.

Keit 121.221.74.211 (talk) 15:17, 21 November 2012 (UTC)[reply]

I wonder how common eight-horse teams were though. Two or four-horse teams seem to have been more often used. Perhaps some advertising sleight-of-hand here. Rmhermen (talk) 21:03, 21 November 2012 (UTC)[reply]

One should always look out for some promotional sleight-of-hand. However, an 8-horse team can apparently pull the same equipment as the tractor, a two horse team cannot. An 8-horse team should be cheaper per acre to operate as it still needs only one human driver, but will do 4 times the work. To compare the economics of two-horse teams you need to allow wages for 4 drivers. Ford allowed for 0.5 hour per day to maintain the tractor. I don't have experince with the 1916 Model F, but I have experience with a restored Model N (quite similar), and 0.5 hour per day is excessive. It takes only 10 minutes to refuel from drums and add coolant (they use quite a bit) and 5 minutes (after you've done it a few times) to do daily checks. So, it looks like Ford were being very conservative. It doesn't generally pay to bend the data for things made for commercial purposes. If a manufacturer tells a businessman a machine costs $10/hour to run, and it actually costs $11/hour, that may affect the viability of the business. That in turn will give the manufacturer a bad name, and could result in a demand for a refund or even a lawsuit. But if a farmer finds the tractor outperfoms the manufacturer's data, he'll tell his friends - that's worth more than any amount of advertising. Keit 60.230.223.27 (talk) 00:01, 22 November 2012 (UTC)[reply]

I also, now that I think about it, have an engineering textbook that gives data on the thermodynamic efficiency of animals (ie calorific value of food eaten vs mechanical power output) - I will post again later. You could compare this with the calorific value of crops converted to engine fuel and the efficiency of engines (typically ~40% for diesel engines) Keit 121.221.74.211 (talk) 15:55, 21 November 2012 (UTC)[reply]

Googling fuel efficiency of horse returns quite a bit of interesting stuff. It seems that the fuel calorific value is converted to mechanical effort by well execised draft horses at about 15% efficiency. What you should perhaps take into account is that the exhaust (horse poo) retains a considerable amount of the calorific value - you can burn it for heat (run a vapor engine to generate electricity??). Apparently muscles have a chemical-to-mechanical conversion efficiency of well over 25%, but there are losses to maintain the rest of the animal, and a horse's ability to digest out all the calorific value of its food is not very good. Ratbone 60.228.252.120 (talk) 00:43, 22 November 2012 (UTC)[reply]