November 27 Information

amino acids > peptides >protein 93.126.88.30 ( talk) 04:35, 27 November 2016 (UTC) reply

See Protein and Peptide. It's more correct to say that a protein is a large collection of amino acids and a peptide is a small collection - the boundary between the two is arbitrarily set at 50, but there's no real biological basis for this number. Proteins can be artificially formed from peptides (see peptide synthesis), but in nature they're formed from individual amino acid residues. Tevildo ( talk) 10:22, 27 November 2016 (UTC) reply

My feeling is that proteins and peptides are the same thing, but this is happenstance rather than definition. There was a time when people did not know that proteins were peptides. According to [1], which seems as likely a place to look as any, "protein" was first studied by Antoine Fourcroy and described much later by Gerhardus Johannes Mulder. The first guy was a contemporary of Lavoisier, who named stuff like hydrogen and oxygen, so he didn't have much idea about peptide bonds!

Now where this gets interesting is that tomorrow we could find some bacterium equipped with a trick ribosome that strings boron and phosphorus compounds or fatty acids or any other thing your fervid imagination might desire into the sequence of an otherwise amino acid based protein. Such a thing would be a protein, but not a polypeptide.

For this reason, my call is that properly a "peptide" is any molecule you know is all peptide - you've synthesized it, or you've nailed down every non-H atom on the crystal structure, and you actually know you can make it entirely out of NH2-C(R)-COOH units. (Note many crystal structures have 'disordered segments', though based on the genetic code you can make a very good guess what they are) Until proven otherwise, a new band on a gel is... well, it's very likely a peptide, but let's call it a protein. Technically I could argue that even something like myristoylation disqualifies a protein as a peptide per se, but the way that chemistry works is that you can simply prepend the qualifier, so you can call it a "myristoylated peptide" anyway. That said, you should probably just go with what Tevildo said. Wnt ( talk) 14:24, 27 November 2016 (UTC) reply

Don't polypeptides have to be folded the correct way to be described as a protein? LongHairedFop ( talk) 19:55, 27 November 2016 (UTC) reply

Well, a protein can be denatured and sometimes even renatured, and by that standard, I'd say denatured protein is protein. Wnt ( talk) 02:56, 28 November 2016 (UTC) reply

108.252.141.219 ( talk) 05:04, 27 November 2016 (UTC) reply

I searched for "terraforming" in the archives of the journal Icarus, and found:

• Terraforming: Engineering Planetary Environments (1997)
• Terraforming Mars: A review of current research (1998)

...so it looks like even Icarus hasn't published much on the topic in a while. If reputable and legitimate scientists believed that new information might have developed in the last few years, there'd probably be something in that journal.

A few years ago, at the 2013 meeting of the American Geophysical Union, a special session was devoted to Geoengineering: "deliberately manipulating physical, chemical, or biological aspects of the Earth system." At least a few responsible scientists believe that our species now possesses the technology to actually induce planetary-scale effects; many believe in anthropogenic climate change; and some scientists believe that with adequate planning, we could use the same technology to plan and produce desirable planetary-scale changes - all using existing technology. Perhaps an even smaller number of scientists believe we might use that technology on worlds other-than-Earth.

But for the most part, terraforming - of Earth, of Mars, of our Moon, or of any other world - is still science fiction. We might, some day in the future, develop newer technologies to harness and control massive quantities of matter and energy - but today, the effects of our actions are usually too small to alter an entire world; and when our technology does have world-altering side-effects, those effects are not the result of a deliberate plan. So - no terraforming of the moon is likely, at least not during the next few years.

Here is some excellent further reading: Why We Explore, a series from NASA's office of the Chief Historian; and a book cited therein: The Railroad and the Space Program. An Exploration in Historical Analogy, (1965). See if you can track down a copy at your local library or college.

Nimur ( talk) 06:49, 27 November 2016 (UTC) reply

Terraforming of the Moon is our redirect on the subject, but the target fails to mention the possibility or otherwise. Graeme Bartlett ( talk) 10:39, 27 November 2016 (UTC) reply

It doesn't seem practical, since the gravity there is too little to keep an atmosphere from escaping into space, due to the solar wind and tidal forces from the Earth, even if you could manage to create one. (You could create habitable domed or underground areas, but that's not really terraforming.) StuRat ( talk) 18:26, 27 November 2016 (UTC) reply

I read once that if the Moon were somehow given an Earthlike atmosphere it would lose one-tenth in 1e4 years. — Tamfang ( talk) 00:33, 28 November 2016 (UTC) reply

That process is Jeans' escape - and in many planetary physics classes, it's a common homework problem to compute the time-constant for many familiar moons and planets. The equation governing this time scale is an equilibrium (or non-equilibrium) between gas diffusion and gravity. More complicated versions account for thermal effects, solar wind, creation of new gas via geological processes, and so on. Nimur ( talk) 02:36, 28 November 2016 (UTC) reply

Keep in mind that terraforming is expected to take thousands of years, so even a slow bleed-off rate is problematic over such a long period. The option of building domes or tunnels to contain the atmosphere might be a good first step, with the idea of building up atmosphere in those places for eventual release into the atmosphere, once we have a solution to the bleed-off. StuRat ( talk) 17:22, 28 November 2016 (UTC) reply

• No, you can't terraform the Moon as such, it is too small to hold an earthlike atmosphere with its gravity (ec with Tamfang). Colliding Venus and then Ganymede with Mars will give a body with a mass slightly less than the Earth's, and a lot of water, but sneaking Venus past Earth would be a delicate operation. You'd also have to wait perhaps tens of millions of years for the surface to cool and stabilize. In the meantime, you could tow in a body like Titan to make a cool moon, since tides are vital in maintaining productive coastal ecosystems. Finding an Earthlike exoplanet would probably be a lot easier and quicker. μηδείς ( talk) 00:42, 28 November 2016 (UTC) reply

^{citation needed}? Did you just make this stuff up, or can you attribute those ideas and concepts to an actual scientific reference? Nimur ( talk) 02:38, 28 November 2016 (UTC) reply

I dunno, Nimur. I took physics for science majors and passed, and looked up the masses of the bodies I mentioned. If you have a specific question, ask it, and identify which of the many people here who have said that the moon won't hold an atmosphere whom you are addressing. μηδείς ( talk) 04:17, 28 November 2016 (UTC) reply

• Yes you can, given sufficiently advanced technology. However, if you want it to last more than 10,000 years or so, you will need to actively maintain your atmosphere, which will otherwise gradually bleed into space. I think you could reasonably expect to find a technical fix for this little problem during that 10,000 year window, but I suspect your goals will have changed by then. - Arch dude ( talk) 00:58, 28 November 2016 (UTC) reply

• Might the relatively low amount of geological activity within the Moon also be a problem? ← Baseball Bugs ^{What's up, Doc?} carrots→ 02:23, 28 November 2016 (UTC) reply

Is any Moon soil like ground glass? You might want to avoid running barefoot through such areas. Sagittarian Milky Way ( talk) 02:47, 28 November 2016 (UTC) reply

If you're going barefoot on the airless Moon, getting cuts on your feet might be the least of your worries. ← Baseball Bugs ^{What's up, Doc?} carrots→ 15:20, 28 November 2016 (UTC) reply

I mean after there's air and an ozone layer. Sagittarian Milky Way ( talk) 16:31, 28 November 2016 (UTC) reply

Some obvious challenges include:

• Getting suitable atmosphere material to the Moon - see Interplanetary Transport Network, perhaps.
• Keeping the atmosphere in long-term - wrapping it in plastic seems like an option, but maybe you can think of something fancier.
• Spinning up the rotation rate to deal with hot and cold extremes - requires a lot of energy/propellant/time, you'd think.
• Dealing with tectonic activity from the spin-up and renewed tides - time and earthquake prediction, I suppose.
• Given the time needed for some other steps, I think the beaches will churn themselves to proper sand in the meanwhile.

Benefits include the having the best orbit around the best star, the best geological materials a planet can have, not to mention the human powered sport flying and the truly massive surfing opportunities. Wnt ( talk) 03:18, 28 November 2016 (UTC) reply

I just read our entire article on the Interplanetary Transport Network... it's not outright wrong, but it presents a very over-simplified description that might lead the unskilled astrophysicist into some very very wrong conceptual conclusions. Although lines of equipotential energy connect many places in our solar system, it is not valid to suggest we can use "zero fuel" or "zero net change in energy" to get from point A to point B along these lines. The key insight that is missing is that, in orbital mechanics, energy takes the form of stored gravitational potential energy and also kinetic energy. In one simple example: your space ship might get to a distant point, but at the wrong velocity to stay at that point! The only way to get to, and then stay at, the point where you want to be, when you're orbiting, is to change your kinetic energy - delta v - which means, using the only technology we have today, to expend fuel. If we forget about this "small detail," when reading our Interplanetary Transport Network article, we reach the wrong conclusion based on a very superficial understanding of the physics. We might conclude that we can use these complicated orbits to "cheaply" and "efficiently" navigate around, following equipotential lines and following unstable orbits without firing the rockets. That's the wrong conclusion: we can pass by a lot of points, but we won't stay there... unless we pursue massive rockets for very large delta v; or pursue direct entry and slam into the target at orbital velocities - slowing from orbit by colliding with the destination! In both cases, we suffer the same issue: today's technology does not enable us to apply this orbital approach to move non-trivial quantities of cargo. We can't build giant rockets cheaply and reliably enough to provide that much delta-v at the destination; and we don't have great technology options for atmospheric entry when the entry-velocity is extraordinarily large!

A great book, for the interested reader, is To Rise From Earth, available from many book resellers. It presents orbital mechanics in a format that a non-mathematician can appreciate, and lays a great foundation for the future physicist who will (eventually) study the equations and mathematical formalisms that describe spaceflight. Another fantastic book is the much-harder-to-find Realities of Space Travel. Our technology has changed since it was authored, but the harsh laws of gravity have not - and despite our best efforts, we really have not learned much to change our understanding of the basic physics of gravity in over a century.

Nimur ( talk) 03:53, 28 November 2016 (UTC) reply

One does need a lot of energy to get out of the gravity well around planets, but otherwise Gravity assist does enable travel around with expending much energy. That's how Voyager manged to get out of the Suns gravity well. One might have to wait for a fruitful conjunction but the energy is got from speeding or slowing the planets by imperceptible amounts. Dmcq ( talk) 13:28, 28 November 2016 (UTC) reply

Gravity assist is great and useful, but to use the technical terminology, it is only able to perform conservative work. If you actually want to change your orbit, you need delta v, which requires non-conservative work. To use non-technical terminology, that means either hitting a planet (crushing on its surface or aerobraking in its atmosphere); or, expending rocket fuel. There are no other options: every option is a variation on these themes. Using gravity assist, one can expend fuel or impart a collision at the best possible time - but there is no way to steal energy from the planet's orbit unless you collide with the planet. Complex equations and rotating reference frames do not invalidate conservation of energy.

Don't let the complications fool you - orbital mechanics is really built on very basic physical law: gravity is a conservative force, so it requires exactly the same amount of energy to get to your destination no matter what path you use. Gravity-assisted orbit boost, or even orbit-capture, does not change this fundamental principle. All it does is permit the engineers to move around when the spacecraft's stored energy gets spent, so that it gets spent when it is the best and most efficient time and place. Gravity-assisted direct-entry (implying a collision with the planet's atmosphere or hard surface) is another really neat option - but with today's materials technology, our spacecrafts cannot survive entry, descent, and landing, from interplanetary orbit velocity. In non-technical terms, the spacecraft will break up or burn up in the planetary atmosphere; or crush on the planet's hard surface. If we can engineer materials and structures that could withstand that event - say, some amazing foamed-titanium truss structure that could crush and dissipate the energy from an impact at 20 kilometers-per-second - as recommended in this 1960 NASA technical report, Landing Energy Dissipation for Manned Reentry Vehicles, we might be able to land somewhere without firing rockets. Unfortunately, NASA scientists realized that no available material in the 1960s could yield a survivable landing - which is exactly why our Apollo missions used giant rockets to perform orbital insertion and landing on the moon; and more giant rockets to perform Earth orbit reentry, followed by aerobraking, followed by parachutes, followed by a just-barely-survivable-by-human water landing. Aerobraking is not a great option on many worlds we might want to visit, especially in places where the atmosphere is sparse - so it's either crushable landing gear, or giant rockets.

Nimur ( talk) 15:35, 28 November 2016 (UTC) reply

What you say is wrong an has been proven wrong in practice. It would have required a huge rocket to get Voyager away otherwise. If one only had a satellite and a single planet then one would not be able to steal energy just from the planet it set off from, but it is a Three-body problem effect going around various planets. Getting off a planet or down to one is not helped by gravity assist, only moving around once one can get off enough to visit another body. Dmcq ( talk) 15:49, 28 November 2016 (UTC) reply

Voyager did not stay, and could not stay, at the planets it visited. It passed beyond them. This is a very important detail. The Voyagers were launched on some of the most powerful rockets ever built; a lot of kinetic energy was imparted to the space probes when they left Earth. The probes traveled to the outer planets along an orbital trajectory. For Voyager to stay at any of the planets it passed by would have required a change in its orbital trajectory, which means an even bigger rocket. Notice that neither Voyager ever entered any planetary orbit - it was never captured (and could never be captured) during any of its gravitational slingshot maneuvers. Compare that spacecraft to, say, the Cassini mission, and look how different the spacecrafts looked. Cassini entered planetary orbit with a huge expenditure of fuel from a massive chemical rocket. Nimur ( talk) 16:37, 28 November 2016 (UTC) reply

As I said Gravity assist doesn't help with taking off or landing on a planet, there is no intermediary body to gain or lose energy from when in its gravitational well. But once one can reach other bodies it is most certainly possible to gain or lose energy. What helped Voyager leave the solar system was gravity assist, it did not have enough speed on launch to escape the sun. The rockets were large for Voyager to take it to Jupiter directly instead of using gravity assist from the inner planets. New Horizons, the fastest rocket ever made, is still slower than Voyager. When going to Mercury gravity assist is used to slow down satellites so they approach it much slower than they would otherwise - the speed is slowed such that if they crashed on the planet it would be with something close to the escape velocity for Mercury. Dmcq ( talk) 17:44, 28 November 2016 (UTC) reply

@ Nimur: See [2] - "a low-energy transfer to a lunar-libration orbit saves 400 meters per second (m/s) of deltaV and often more. This is a significant savings..." I understand this network isn't quite magic, but the three body problem notoriously interjects some rocket science into rocket science, I mean, you can't just calculate everything in a neat little equation, it takes modelling and clever thinking - cleverer than I claim to bring to the table, to be sure. Wnt ( talk) 00:42, 29 November 2016 (UTC) reply

Right - 400 m/s less than a less-efficient orbital transfer! The objective, in orbital trajectory planning, is usually to work towards the theoretical best orbital trajectory, or minimum possible delta-v. There's no way to change your orbit with zero delta-v; but if you're not at the theoretical minimum, you can find ways to use less delta-v. Nimur ( talk) 19:45, 29 November 2016 (UTC) reply

I don't want to defend a strawman here - I realize you can't get off the mountaintop with clever orbital engineering. But see [3] page 3:

In general, a low-energy transfer is a nearly ballistic transfer between the Earth and the Moon that takes advantage of the Sun’s gravity to reduce the spacecraft’s fuel requirements. The only maneuvers required are typical statistical maneuvers needed to clean up launch vehicle injection errors and small deterministic maneuvers to target specific mission features. A spacecraft launched on a low-energy lunar transfer travels beyond the orbit of the Moon, far enough from the Earth and Moon to permit the gravity of the Sun to significantly raise the spacecraft’s energy. The spacecraft remains beyond the Moon’s orbit for 2–4 months while its perigee radius rises. The spacecraft’s perigee radius typically rises as high as the Moon’s orbit, permitting the spacecraft to encounter the Moon on a nearly tangential trajectory. This trajectory has a very low velocity relative to the Moon: in some cases the spacecraft’s two-body energy will even be negative as it approaches the Moon, without having performed any maneuver whatsoever.

There is an interesting map of Jacobi constants that determine "forbidden regions" by velocity on page 38. I think but certainly will not swear that provides some hard limits. I am not sure if this is related to the statement on page 13 that "Sweetser computed that the theoretical minimum ΔV that a space­ craft would require to travel from a 167-km altitude circular orbit at the Earth to a 100-km altitude circular orbit at the Moon, just passing through L1, is approximately 3.721 km/s" It is fairly clear from the text and the complicated shape of these bands that there are some transfers possible within a "network" that require no energy, once a spacecraft has gotten to some location. Wnt ( talk) 12:48, 30 November 2016 (UTC) reply

• An interesting alternative: [ https://blogs.scientificamerican.com/guest-blog/lets-colonize-titan/ ] -- Guy Macon ( talk) 17:15, 28 November 2016 (UTC) reply

Oh, nonsense. "Titan is the only place..." I mean, look up from that frigid wasteland to Saturn. It's got the perfect gravity, liquid water, with a few percent oxygen added you can make hydreliox mix to live at 10 atm where that is, and there are bands of relatively calm, upwelling atmosphere. True, you have to fly all the time and a shaker of salt is probably more precious than gold on Earth, and yeah the next layer of atmosphere is industrial grade stink bomb, but relatively speaking, doable. Venus is probably better, though. Wnt ( talk) 00:48, 29 November 2016 (UTC) reply

...Ultimately, what will be the distribution? Answer: It will be equally likely to find any pair moving in any direction in space. After that further collisions could not change the distribution.

They are equally likely to go in all directions, but how do we say that? There is of course no likelihood that they will go in any specific direction, because a specific direction is too exact, so we have to talk about per unit “something.” The idea is that any area on a sphere centered at a collision point will have just as many molecules going through it as go through any other equal area on the sphere. So the result of the collisions will be to distribute the directions so that equal areas on a sphere will have equal probabilities.

Incidentally, if we just want to discuss the original direction and some other direction an angle θ from it, it is an interesting property that the differential area of a sphere of unit radius is sinθdθ times 2π (see Fig. 32–1). And sinθdθ is the same as the differential of −cosθ. So what it means is that the cosine of the angle θ between any two directions is equally likely to be anything from −1 to +1.
— Feynman • Leighton • Sands, The Feynman Lectures on Physics, Volume I

Feynman first explains that directions of molecules are random, but probability that some molecule go through some area on the imaginary sphere is equal for equal areas. I agree. Last paragraph is unclear.

According to Fig. 32–1 ${\displaystyle dA=2\pi r^{2}\sin \theta \,d\theta }$. dA is a change of the area between angle θ and (θ+dθ). But this area is the area of spherical segment limited by 2 conical surfaces, not 2 directions, and not 2 equal areas.

Feynman suggests next manipulation ${\displaystyle dA=2\pi r^{2}\,d(-\cos \theta )}$
and for r=1 ${\displaystyle dA\propto -d\cos \theta }$ ;

${\displaystyle A_{12}\propto \int _{\theta _{1}}^{\theta _{2}}-1d\cos \theta =\cos {\theta _{1}}-\cos {\theta _{2}}}$

If e.g. ${\displaystyle \cos {\theta _{1}}-\cos {\theta _{2}}=\cos {\theta _{5}}-\cos {\theta _{6}}}$, then ${\displaystyle A_{12}=A_{56}}$ and then these areas have equal probabilities. But I can't see it's correct for cosine itself. Besides different dA's are different because ΔCosθ is different for different angles (maximum at θ = π/2).

Can you show how Feynman derived that cosine must be from −1 to +1?

Username160611000000 ( talk) 08:08, 27 November 2016 (UTC) reply

This seems like a long and tortuous way of saying the result by Archimedes that the area of a slice of a sphere is the same as the area sliced on a cylinder round the sphere but I suppose trigonometry and calculus are the way to do that now. Anyway see [5] Dmcq ( talk) 11:16, 27 November 2016 (UTC) reply

Maybe Archimedes' Hat-Box Theorem is correct for equal h (or dh), but we use equal dθ. Username160611000000 ( talk) 11:25, 27 November 2016 (UTC) reply

Even if ΔCosθ would be equal with increasing θ on equal steps, I still don't understand the logic. Why does cosine of the angle between 2 directions of 2 pairs of molecules vary from -1 to 1? This statement is absolutely not correlated with previous one about differential area . Username160611000000 ( talk) 11:35, 27 November 2016 (UTC) reply

Feynman next uses this statement to prove that average cosine is zero. So I need to understand this paragraph by all means. — Preceding unsigned comment added by Username160611000000 ( talk • contribs) 11:39, 27 November 2016 (UTC) reply

I assume that a conic surface is associated with some direction of the spread of pair of molecules (the spread of 2 molecules occurs in opposite directions). So molecules can go in any direction inside surface. Then this conic surface is associated with little area (in fact the section of surface and sphere (circle) must be in the middle of the area , but with infinitesimal values it's not important). Is it correct? Username160611000000 ( talk) 11:58, 27 November 2016 (UTC) reply

The cosine of θ is −1 in the straight downwards direction and +1 in the straight upwards direction. Dmcq ( talk) 12:53, 27 November 2016 (UTC) reply

I agree. But it doesn't mean that every direction is equiprobable, because it does not connected with equal areas. Username160611000000 ( talk) 13:09, 27 November 2016 (UTC) reply

They said equal areas were equiprobable. They never said that equal changes in θ gave equal areas. Dmcq ( talk) 13:24, 27 November 2016 (UTC) reply

dθ is used to calculate differential area , every dθ is equal. Why is this differential area needed at all ?

To show that angle can get any value between 0 and π with equal probability, we must show that there is some area associated with each direction in some way and that these areas are equal. Feynman did not show that, he just said that there is some differential area. So what? Differential area exists. It is not connected with direction, it is not connected with cosine. Username160611000000 ( talk) 13:54, 27 November 2016 (UTC) reply

To get equal differential areas of segments you must have equal d(cos θ) not dθ. Nobody put in any requirement for every differential of θ to be the same. When you have dy/dx you get the ratio of changes out and you do not expect a constant. Dmcq ( talk) 14:15, 27 November 2016 (UTC) reply

When you have dy/dx you get the ratio of changes out and you do not expect a constant. When we build a graph of g(x)=dy/dx we take some point y(1) and find Δy during some Δx, then we take next point y(2) and again find Δy during SAME Δx. 1st and 2nd Δx are equal, 1st and 2nd Δy are not equal. Username160611000000 ( talk) 14:27, 27 November 2016 (UTC) reply

To get equal differential areas of segments you must have equal d(cos θ) Agree. What are the next manipulations with d(cos θ)?

No the Δx do not have to be equal. d(cos θ) is equal for equal displacements in the vertical direction. Dmcq ( talk) 14:41, 27 November 2016 (UTC) reply

d(cos θ) is equal for equal displacements in the vertical direction. Ok. Let displacements be equal. What is next? Username160611000000 ( talk) 15:04, 27 November 2016 (UTC) reply

Let the vertical diameter of the sphere consist of ten parts Δh₁, Δh₂ ..., Δh₁₀. Consider some vertical displacement Δh₃. What angle θ is associated with Δh₃? θ ∈ (53°;66°], probability P(θ ∈ (53°;66°])= 10%. Consider Δh₂, θ ∈ (36°;53°], probability P( θ ∈ (36°;53°])= 10%. What is probability θ ∈ (53°;54°]? It is ${\displaystyle P(\theta \in (53^{\circ };54^{\circ }])={\tfrac {P(\theta \in (53^{\circ };66^{\circ }])}{13}}={\tfrac {10\%}{13}}=0.8\%}$.
By analogy
${\displaystyle P(\theta \in (36^{\circ };37^{\circ }])={\tfrac {P(\theta \in (36^{\circ };53^{\circ }])}{17}}={\tfrac {10\%}{17}}=0.6\%}$;
${\displaystyle P(\theta \in [0^{\circ };1^{\circ }])={\tfrac {P(\theta \in [0^{\circ };36^{\circ }])}{36}}={\tfrac {10\%}{36}}=0.3\%}$.
So we get increasing probabilities in equator direction [ png] . Username160611000000 ( talk) 15:12, 27 November 2016 (UTC) reply

Diapason is not a term normally applied in anything like this. It is a musical term and only applied in other contexts to mean a range in a literary sense. Range would be a better name and range does not apply to a single angle. Dmcq ( talk) 13:08, 28 November 2016 (UTC) reply

Also formula for A is A(θ)=2πR²(1-Cosθ). Letting R=1 and expressing θ we get : θ(A)=arccos(1-A/2π). ${\displaystyle {\tfrac {d\theta }{dA}}={\tfrac {d}{dA}}\arccos {(1-{\tfrac {A}{2\pi }})}={\tfrac {1}{2\pi {\sqrt {1-{(1-{\tfrac {A}{2\pi }})}^{2}}}}}}$. graph . We see the angle changes nonlinearly. Username160611000000 ( talk) 17:04, 28 November 2016 (UTC) reply

The relevance is fairly marginal, but I went through the same kind of confusion looking at the distribution of the sine of an angle (which is not uniform) in this previous discussion here. Wnt ( talk) 14:39, 27 November 2016 (UTC) reply

It is correct that the area that Feynman talks about here is the area of spherical segment limited by 2 conical surfaces, and they are not equal areas. So at the smaller differential areas (nearer to the 'poles') there would be a smaller number of molecules going through than compared to larger differential areas (nearer to the 'equator'). Feynman is talking about the relationship between the number of molecules passing and θ. Imagine a circle, with along the circle marked the values of the cosines the line through that point and the center makes with a fixed axis. Now mark off tiny segments along the circle, such that the difference of the cosines at the ends of each segment are equal. So these segments mark off equal d(cos θ). Our circle here can be the 'profile view' of the sphere that Feynman was describing. So, through each of the segment there is an equal amount of molecules passing, as they correspond to equal areas on the sphere. A randomly picked molecule could be passing through any of our segments equally well. But then each segment has divided cos θ equally, as per our construction. Hence, 'the cosine of the angle θ between any two directions is equally likely to be anything from −1 to +1'. Questintelligence ( talk) 11:08, 26 December 2021 (UTC) reply

what are proteins, lipids, nucleic acids and carbohydrate classified? I mean to ask if they are groups of or families of biochemistry. For example I want to say that we have 4 families in biochemistry: proteins, lipids, carbohydrates, and nucleic acid. Is it true to refer them like that? 93.126.88.30 ( talk) 15:46, 27 November 2016 (UTC) reply

Generally speaking, chemicals involved in biology ("biochemistry") are described as belonging to four families, amino acids (including proteins), lipids, carbohydrates, and nucleic acids (including nucleobases, nucleosides, DNA, RNA, etc.). That can lead to some amount of oversimplification, however, as it doesn't necessarily account for some more complex compounds, such as chlorophyll or heme porphyrins with metal centers, some vitamins, etc. -- OuroborosCobra ( talk) 19:36, 27 November 2016 (UTC) reply

I took the question as asking for a category that includes the named groups but excludes others. You can generalize, using such terms as organic molecules, but that includes things like octane and methyl alcohol, both dangerous toxins. You could mention nutrients, but not all carbohydrates, for example, are nutritious, and we don't require nucleic acids as such, since we can synthesize them. This sounds like a homework question. Is it? I have a suggestion, but the poster should tell us what he thinks before we supply suggestions. μηδείς ( talk) 00:29, 28 November 2016 (UTC) reply

One clue is to see how Wikipedia classifies them. If you read the articles, down the bottom will be categories. You can click on these categories and see what else is in them, or what their parent categories are. Graeme Bartlett ( talk) 05:43, 28 November 2016 (UTC) reply

Commons also has a category tree eg Commons:Category:Lipids. Graeme Bartlett ( talk) 11:42, 28 November 2016 (UTC) reply

• This page and These lecture notes basically agree with the above, regarding the four major classes of biomolecules. Wikipedia's article titled Biomolecule doesn't directly cover the 4-class structure, but does also go into other types of biomolecules that cannot be classified such as terpenes, steroids, and lignin, etc. -- Jayron 32 14:11, 28 November 2016 (UTC) reply

Biochemistry#Biomolecules gives the same four major types, cited to a published textbook. DMacks ( talk) 14:16, 28 November 2016 (UTC) reply