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August 19

Pressure increase on the ground due to an overflying plane

A plane of mass m flies at some altitude h (say h = 10 km) above the ground. The force of gravity acting on the plane is ultimately transferred to the ground, so the pressure at the ground will increase due to an overflying plane and this obviously depends on the position on the ground relative to the plane. The problem is then to compute this pressure increase from first principles. You can only simplify things by assuming that the plane is very far away from the ground. Count Iblis (talk) 03:40, 19 August 2012 (UTC)[reply]

Sounds like homework. The problem is not knowing how the pressure is distributed. If forced to guess, I'd go with a 45 degree cone with equal pressure applied to all the ground under it. Has anyone actually studied this to determine the true force distribution ? StuRat (talk) 04:04, 19 August 2012 (UTC)[reply]

The pressure quickly spreads out spherically. The mean free path of an air molecule, which moves at 1,000 km/h at STP is 10 cm. This conical notion seems to imply you are forgetting air is a gas. μηδείς (talk) 04:12, 19 August 2012 (UTC)[reply]

Spherically ? Do you mean hemispherically ? Certainly a plane doesn't increase the pressure above it. StuRat (talk) 04:15, 19 August 2012 (UTC)[reply]

Yes, it does, there is nothing to constrain it. Consider a sonic boom. It doesn't just travel downward or in one direction. When you blow up a balloon, it does not expand only on the surface opposite your mouth. You are imagining a gas as if it were a pile of sand with a critical angle of repose. The atmosphere consists of particles bouncing off each other in random directions at 1,000 km/h at STP. Any change in pressure (and actually there is no real net change in pressure except the from the heat generated by the flight, since the difference in pressure above and below the wing is quickly neutralized as the plane passes) is rapidly transmitted in every direction. μηδείς (talk) 04:27, 19 August 2012 (UTC)[reply]

It can be useful to think of the atmosphere as an ocean of air. Airplanes, especially those flying at high speeds create a "wake" in the air just as boats do on a pond. And if another traveling object hits that wake, it can cause it to "bounce" in the unstable substance. ←Baseball Bugs ^{What's up, Doc?} carrots→ 05:00, 19 August 2012 (UTC)[reply]

Yes, but the wake (of turbulence, not pressure) soon dissipates, and is not a cone of pressure beneath the airplane. μηδείς (talk) 05:06, 19 August 2012 (UTC)[reply]

I think I see the difference. However, I've been on planes when we got bounced around a bit, and the pilot said we were in the wake of another plane. ←Baseball Bugs ^{What's up, Doc?} carrots→ 05:52, 19 August 2012 (UTC)[reply]

Yes, wake turbulence is a very real thing, but it is different to the equal and opposite reaction to the lift, which is what the OP is asking about. --Tango (talk) 10:45, 19 August 2012 (UTC)[reply]

Fascinating question with very real implications, and very difficult physics to contemplate. Every time I land my little Citabria at San Jose International Airport, the tower warns me to watch out for vortices and wake turbulence from the big 737s on every single approach - even if I'm the only aircraft around - just in case the vortex from the last jet is still "floating around" in the air. The textbook-solution (Chapter 7 section 3 of the AIM) has some nice cartoon diagrams of persistent wake and vortex effect. 7-3-4 discusses sink-rate of persistent vortices - in other words, how long until the wake "hits" the ground. So, that's how to solve it not from first principles, but in a way that provides pilots with useful information.

From first-principles, the assumption is invalid: air is compressible, so the dynamic pressure that creates lift near the aircraft need not ever propagate to the ground. It could aerodynamically convert to another form of energy, such as compression-heating. After the aircraft passes, it could return to its original state. Energy is conserved, and momentum is conserved, but there is no law of conservation of pressure, temperature, or force. In an ideal, perfect, skin-frictionless system with perfect laminar flow, the air should flow smoothly around the aircraft, and no energy need be lost to propagating a pressure wave away from the aircraft surface.

In the vortex wake turbulence case, the air flow is not propagating as a pressure wave - it is a turbulent, convecting effect. For comparison, the sound of the engines is propagating as a pressure wave, at much higher frequencies. It would be interesting to compare total energy or power in each propagating mode. I suspect more energy is propagating as low-frequency convecting turbulence, rather than as an acoustic pressure wave. NASA has studied wake turbulence and the vortex effect very thoroughly - there's a classic photo that is now a featured picture - aircraft vortex effect. To compute these sorts of things from first principles, as the OP requested, is nearly impossible: but that's how we derive the basic laws of air flow such as Navier Stokes. I think the problem is the assumption that the dynamic air motion can be expressed simply: but in an enormous mass of thermally-inhomogeneous, convecting air, simple pressure-waves are so unphysical as to be misleading. Suffice to say, the aircraft wing loading is sustained by the dynamic pressure of the air close to the aircraft; and that air is supported by the dynamic pressure of the air near it, ... and ad infinitum, all air that ever contacts any other air exchanges energy and momentum through complicated aerodynamic processes. The speed of propagation of each effect is different: there is an effective wave speed for pressure, for shear/vortex propagation, for thermal conduction, and so on. Each process conveys some of the energy, and some of that energy translates into transient, dynamic pressure. Because air is compressible, it's meaningless to talk about only the pressure wave propagating. You must consider the other aerodynamic effects. Nimur (talk) 16:00, 19 August 2012 (UTC)[reply]

Pressure always acts in all directions. That's the main thing that distinguishes between it and other forces. Hold you hand out in front of you, flat to the ground. The mass of the column of air above it is about 8kg and the weight of that is pressing down on your hand. That's the air pressure. You can hold your hand up easily, though, because there is an equal pressure on the bottom of your hand and the forces cancel out. While the pressure on the top of your hand can be thought of as the weight of the air above it, the pressure of the bottom is a little more difficult to understand - it's caused by the rest of the atmosphere pressing down on the air next to your hand, which pushes on the air under your hand, which pushes up on your hand. --Tango (talk) 10:45, 19 August 2012 (UTC)[reply]

The problem with the "cone" model is that the force won't immediately be transmitted to the ground, but will be transmitted as some sort of slow pressure wave. The problem is a bit similar to blowing briefly at the far wall of a room - where will the breath impact? - except that because the plane has since moved the area "supporting" it will be far behind it when it reaches the ground. A difficult and interesting question. Wnt (talk) 14:27, 19 August 2012 (UTC)[reply]

As pointed out above, a plane's downwash spreads outward and dissipates rapidly. When helicopters and planes fly low enough, this downwash interacts with the ground (see the article on ground effect (aircraft) for instance). With helicopters their downwash can create vortex rings of air that are dangerous, because the air mass in which the copter is in starts to descend rapidly. Air vortex cannons can create these vortex rings which can hit targets such as the ground, and these cannons can be as simple as thumping the bottom of an empty one-gallon milk jug. Modocc (talk) 15:19, 19 August 2012 (UTC)[reply]

Spreads outward, yes; dissipates, somewhat, at least in terms of energy - but one thing it cannot do is dispose of its downward momentum, except by impacting something to transfer it to. Wnt (talk) 15:43, 19 August 2012 (UTC)[reply]

The downward momentum is quickly transferred and scattered amongst a large number of molecules though, adding a nearly insignificant downward movement to the air in a much larger surrounding atmosphere. The entire Earth moves too, eventually, but that happens every time I get out of bed. Also, the energy of the downwash is still dissipated, thus when it has no more energy, it no longer transfers momentum to the atmosphere. -Modocc (talk) 16:20, 19 August 2012 (UTC)[reply]

This is related to that old thought experiment/joke of getting a van full of birds over a bridge that won't take the weight by making sure enough of them are flying at any given time. It doesn't work because the lift on the birds from their wings has an equal and opposite force on the floor of the van. For the van, it's a closed space so you know all the force will go to the floor of the van. For a plane, it isn't a closed space so the force can be much more spread out. It has to still exist, though - there are no exceptions to Newton's third law. --Tango (talk) 17:13, 19 August 2012 (UTC)[reply]

Sure it exists, but its certainly important here to observe which objects this force is transferred to and what significant effects it can have with respect to other influences. In this case, with increases in the region affected, there is increasing mass involved thus there is less possible motion or speed. Moreover, one has to consider too that if I am getting out of bed or flying about, no net movement of the Earth or the atmosphere will occur if I have a twin doing the exact same thing, but creating an opposing force. In fact, such an equilibrium state and the observed lack of net influence is typical. Modocc (talk) 17:41, 19 August 2012 (UTC)[reply]

So far as I know, transferring momentum always requires energy, but the amount of energy can get less and less. (Hence the "somewhat" I used above) For example, a photon can carry a quantum of angular momentum no matter how low frequency (low energy) it is, and as far as I know this doesn't set any minimum energy that photons can have. Likewise the momentum of a speeding bullet can be transferred to a stinging bruise behind a bulletproof vest, to the relatively mild force the person thus shot exerts on a wall to maintain his balance. Wnt (talk) 18:37, 19 August 2012 (UTC)[reply]

When the plane's trailing downwash of air pressure has transferred its energy, the downwash seizes to exist. Its mass-energy and momentum is therefore conserved by the more massive mass-energy and momentum of the atmosphere and the Earth. --Modocc (talk) 19:17, 19 August 2012 (UTC)[reply]

Yes, that "chickens in a sealed truck" thought experiment is an applicable one. Let's extend that by imagining a truck which is miles long, wide and high, with an airplane inside. Does the total system become lighter when the plane is in flight ? Obviously not. Therefore, there must be some downward force to provide the equivalent of the weight of the plane, at those times. I suggest that this is caused by higher pressure on the floor of the truck than on the roof. This, however, contradicts some of the statements above, such as the pressure dissipating evenly (above and below), or changing into heat. StuRat (talk) 20:44, 19 August 2012 (UTC)[reply]

I believe what was meant above was that the plane's contribution to the pressure on the ground, which is force per area, is not going to be significant because it is spread over a wide area. -Modocc (talk) 21:39, 19 August 2012 (UTC)[reply]

All the atmosphere's pressure is eventually transfered to the earth's surface, over the entire earth. Since the atmosphere has no lid, it is simply the atmosphere's total weight which presses on the earth. But air pressure is not the same as the coherent directional movement of air. Adding a plane of a certain weight to the atmosphere is just like adding an equivalent weight of air to the atmosphere. If a plane is actively flying, it is exerting a downward force on the air below its wings which may translate to the ground as wind, if it is close enough. But an airplane within a giant space will simply create circulating winds that will have no coherent direct action on the surface below it.

Any increase in air pressure within a sealed container will exert just as much lifting force on the roof as it will depressing force on the floor. An airplane within an infinitely giant balloon would cause the balloon to expand, not to sink. Sustained flight within too small an enclosed area is simply not possible because the air will become turbulent and circulating currents will become harder and harder against which to generate lift. Within an infinitely large area you are dealing with just pressure--on the earth this is countered eventually by downward force--in an enclosed area it is countered by balanced increasing pressure on all surfaces in all directions. μηδείς (talk) 22:44, 19 August 2012 (UTC)[reply]

For the most part I agree with you, because being trapped in a balloon, a bee or plane won't alter the balloon's buoyancy which is fixed, but the internal weight distribution can change with flight. Thus, when a plane takes off, it adds its weight to the atmosphere, the contribution of this force (its mass*g) being small... and if its within a giant balloon the balloon expands... which is from an increase in air pressure from the additional energy of the plane's propellants. When the plane lands, its mass-energy is correspondingly lighter and the atmosphere heavier. Its not all that difficult to hover within fairly confined spaces (think of a nearly buoyant swimmer treading water by pushing it downward) and wings produce pressures and partial pressures creating the winds of the downwash that, in turn, produce pressures, like sound waves and pressures against adjacent air masses as well as obstacles, but the essential impulse is usually rapidly absorbed by the surrounding large volume of air in a relatively short distance. Thus, the question of an atmospheric pressure change due to either the weight of the plane or due to the weight of the absorbed energy from a great altitude... seems to be similar to asking how much does the salinity of an ocean basin differs if one adds a grain of salt and ignores all the other contributions to its saltiness. That said, perhaps some tangible numbers, as the OP has requested, would clarify that this is indeed the case. -Modocc (talk) 03:58, 20 August 2012 (UTC)[reply]

Suppose the plane flies around for a long time near the north pole. The gravity exerted by the plane will cause the Earth to be attracted to it with the same force as it attracts the plane, and over time, the Earth will fall constantly northward, accumulating a great deal of momentum (even if it is not particularly visible) in that direction --- unless, that is, some force presses back against the Earth with an equivalent force to that gravity. Now that press-back is not a universal inward force on the entire spherical surface of the Earth, because that wouldn't shove the Earth back in one direction away from the plane. It is a force that remains aligned with the position of the plane, transmitting its weight to the Earth somewhere vaguely below it, in such a way as to add up to precisely the right force to keep the Earth from being pulled by its own bootstraps in any particular direction simply because a plane is in the air. Wnt (talk) 04:16, 20 August 2012 (UTC)[reply]

That's about right, for the Earth's momentum is conserved and does not and cannot change, and its precisely the force of the impulse that is involved, for it is absorbed, shoving the Earth and its atmosphere back, including regions above the plane, negating the gravitational pull. Without the Earth's gravity, the impulse would cause the Earth to be pushed southward. Again, an impulse that begins at a high altitude or even a modest altitude, is a comparatively very small force that spreads out over a very large area and becomes essentially absorbed and blended with other more significant interactions, for I think its likely to be a smaller influence than measurable localized barometric changes. We need to pin some numbers on this though. Modocc (talk) 07:01, 20 August 2012 (UTC)[reply]

Ah, aerodynamics. The last remaining branch of classical physics where nobody can provide a coherent answer, but everyone thinks the answers are obvious except those who've actually studied the subject. But yes, ultimately a flying aircraft's weight is transferred to the ground via increased pressure. Take my word on it, I'm not an expert. ;)

As Modocc said, tangible numbers would help here. I'd start by looking into ground effect, where the pressure changes are going to be more obvious. AndyTheGrump (talk) 04:13, 20 August 2012 (UTC)[reply]

Yes, "Ground effect", or just observing a low-flying plane or helicopter from just below, shows the force on the ground. Tango implies that pressure is always the same in all directions, and this is true in a static fluid, but is not true in a moving fluid such as the atmosphere. For a high-flying aircraft, the weight of the machine is ultimately exactly balanced by movement of air over a wide area, transferring an equal and opposite force to the surface of the earth on the side of the aircraft. By the time any pressure changes have been propagated to the far side of the earth, the aircraft will have landed (or completed several subsequent flights). As Andy implies, aerodynamics is very different from static pressure. When RAF planes fly low over my roof, I definitely feel the downward pressure! Dbfirs 11:58, 20 August 2012 (UTC)[reply]

Distance of galaxies from Earth

How many light-years from Earth would a congalaxy have to be if I made it a galaxy that hadn't been discovered by Terran astronomers yet? Subliminable (talk) 05:48, 19 August 2012 (UTC)[reply]

What is a congalaxy? Someguy1221 (talk) 06:01, 19 August 2012 (UTC)[reply]

This link should explain: [1] Subliminable (talk) 06:10, 19 August 2012 (UTC)[reply]

Working in proper lengths, the most distant observed galaxy-candidate is ~30 billion light years away (see Observable universe#Most distant objects). There's an extra ~17 billion light years of radius in the observable universe left to work with. Someguy1221 (talk) 06:24, 19 August 2012 (UTC)[reply]

But what if I wanted to create a galaxy that was 600 million light years away? Would there be a realistic chance that science hadn't discovered such a galaxy yet as of 2012? Subliminable (talk) 06:29, 19 August 2012 (UTC)[reply]

There are (estimated) over 100 billion galaxies in the observable universe, and the vast vast majority of them have never been identified. At 600 million light years you are probably safe. Someguy1221 (talk) 06:34, 19 August 2012 (UTC)[reply]

Or you could place it so it couldn't be discovered. It would have to be hidden somehow. It could be behind a closer galaxy, or blocked by something within our own galaxy, like a gas nebula. StuRat (talk) 06:37, 19 August 2012 (UTC)[reply]

On Stu's note, you could put it on the far side of the Milky Way's center, relative to Earth. Although that's not a position it could maintain, it would easily make it unobservable for the relatively brief period of time we've been using telescopes. Someguy1221 (talk) 06:41, 19 August 2012 (UTC)[reply]

We are still discovering galaxies right at our doorstep, e.g. Ursa Major I Dwarf at 330000 light years. Youjust have to make them small enough. --Wrongfilter (talk) 07:32, 19 August 2012 (UTC)[reply]

We discover distant galaxies by picking a very small region of the sky and looking at it very closely. That means most galaxies haven't been discovered because they aren't in one of the regions we've looked at. --Tango (talk) 10:47, 19 August 2012 (UTC)[reply]

Electric Potential

Consider a circular disc of radius $a$ that lies in the xy plane and surface charge density (in cylindrical coordinates $(s,\phi ,z)$ ): $\sigma (s,\phi )={\frac {\alpha }{s}}\cos \phi$ . Calculate the potential at a point $h{\hat {z}}+r{\hat {s}}$ .

This is my calculation:
The electric field is $\int \int _{A}{\frac {\sigma dA}{(h^{2}+(r-s)^{2})^{\frac {3}{2}}}}((r-s)\cos \phi ,(r-s)\sin \phi ,h)$
$=\alpha \int _{0}^{2\pi }\int _{0}^{a}{\frac {dsd\phi }{(h^{2}+(r-s)^{2})^{\frac {3}{2}}}}((r-s)\cos ^{2}\phi ,(r-s)\sin \phi \cos \phi ,h)$
$=({\frac {-\alpha \pi }{(h^{2}+(r-a)^{2})^{\frac {1}{2}}}}+{\frac {\alpha \pi }{h}},0,{\frac {2\pi (a-r)\alpha }{h(h^{2}+(r-a)^{2})^{\frac {1}{2}}}})$ . Now, to calculate the potential: $-\int _{\infty }^{h}{\frac {2\pi (a-r)\alpha }{z(z^{2}+(r-a)^{2})^{\frac {1}{2}}}}dz$ . Widener (talk) 10:51, 19 August 2012 (UTC)[reply]

Do you have a question? -- Scray (talk) 12:07, 19 August 2012 (UTC)[reply]

He wants us to check his method. Plasmic Physics (talk) 12:11, 19 August 2012 (UTC)[reply]

Yeah. Widener (talk) 12:35, 19 August 2012 (UTC)[reply]

My first question is, is the separation vector right, id est

((r-s)\cos \phi ,(r-s)\sin \phi ,h)

because

h{\hat {z}}+r{\hat {s}}

is not even really a single particular vector in

\mathbb {R} ^{3}

is it. That's a bit confusing. Widener (talk) 12:34, 19 August 2012 (UTC)[reply]

You're right, the separation vector is not correct. I don't understand either what is meant by

{\hat {s}}

since the direction of that unit vector depends on the

\phi

coordinate of the point where the potential is being calculated and it seems to me that this coordinate was not given. Dauto (talk) 22:07, 19 August 2012 (UTC)[reply]

Does it help if I mention that

r

is small? Widener (talk) 23:58, 19 August 2012 (UTC)[reply]

Is this stone is natural or human made?

This local people said that their ancestors worshiped this rock. They told this is a natural formation. But after saw the bottom side of the rock I think this is human made. Can any one guess this period? and artificial or natural? To see the bottom side of the rock exactly visit this blog.

http://fromthenmaduraitotenkasi.blogspot.in/2012/08/1.html

http://4.bp.blogspot.com/-ipo2kpQm7Ck/UC-wuyspMsI/AAAAAAAAAF8/JbT9pZqvkTM/s1600/IMG_0067.JPG

--Tenkasi Subramanian (talk) 17:42, 19 August 2012 (UTC)[reply]

It's natural; see balancing rock. μηδείς (talk) 18:17, 19 August 2012 (UTC)[reply]

I already saw this balancing rocks. http://wowpics.in/amazing-pics/top-10-unbelivable-balanced-stone-around-the-world/

But after see the bottom side only I confused. Did u c the bottom side of the rock? see this link?

http://4.bp.blogspot.com/-ipo2kpQm7Ck/UC-wuyspMsI/AAAAAAAAAF8/JbT9pZqvkTM/s1600/IMG_0067.JPG

Can u say what type of bond is made between floor and rock?--Tenkasi Subramanian (talk) 18:23, 19 August 2012 (UTC)[reply]

They're both just rock - by which I mean that the rock is continuous, its not actually balanced. Mikenorton (talk) 18:28, 19 August 2012 (UTC)[reply]

You're saying it's a hoodoo? μηδείς (talk) 22:18, 19 August 2012 (UTC)[reply]

No, there's no hard caprock here, just a layer that's slightly less resistant to erosion, which forms the narrow part of the rock. Still pretty strong rock though (compressive strength at least) to take that load. Mikenorton (talk) 07:16, 20 August 2012 (UTC)[reply]

Looks natural to me but could have been enhanced by people? Really hard to tell without examining it for tool marks etc... PatHadley (talk) 09:50, 20 August 2012 (UTC)[reply]

Ozone Dangers

Is it a fact that just 5% of Ozone if mixed with the atmosphere can cause all life-forms to vanish ? Also is it true that if ozone is cooled up to -200 ^o, C liquidized, and put into a breakable glass ball, if blasted like a grenade can turn a man into glass? 124.253.89.68 (talk) 17:48, 19 August 2012 (UTC)[reply]

Some one told that Ozone is O3 (Oxygen). I think that's not harmful.--Tenkasi Subramanian (talk) 18:29, 19 August 2012 (UTC)[reply]

Please don't answer science desk questions based on what you 'think'. Check your facts:

"Even very low concentrations of ozone can be harmful to the upper respiratory tract and the lungs. The severity of injury depends on both by the concentration of ozone and the duration of exposure. Severe and permanent lung injury or death could result from even a very short-term exposure to relatively low concentrations." [2]

I suggest you both read our Ozone article. As for the 'ozone grenade', where did you see that? AndyTheGrump (talk) 18:34, 19 August 2012 (UTC)[reply]

Yes, as stated above, ozone is harmful if breathed in, but it's also quite helpful in blocking UV light in the ozone layer. So, we want lots of ozone in the air, just not down here where we would breath it. StuRat (talk) 20:24, 19 August 2012 (UTC)[reply]

Just to clarify, we want lots as in ~5-10ppm, not ~5%. 203.27.72.5 (talk) 03:32, 20 August 2012 (UTC)[reply]

Given it's half-life of 30 minutes in the lower atmosphere, I think fish in the sea would be safe and maybe some or most trees would survive. But it's highly corrosive and has a IDLH of 5ppm, making hydrogen cyanide with it's value of 50ppm seem quite benign in comparison. Found some rumours about India’s glass man, with a skeleton visible inside. Liquid ozone can detonate, but to turn a man into glass, you'd need a man made of silica to begin with ... Ssscienccce (talk) 23:23, 19 August 2012 (UTC)[reply]

Glass has more than one meaning, and flash-frozen flesh meets the most general definition. μηδείς (talk) 23:29, 19 August 2012 (UTC)[reply]

I did some back of the envelope calculations using the following data; density of liquid O₃ = 1.352g/mL [3], heat capacity of liquid O₃ ~ 0.45cal/gK[4], ozone boils at 161K, the latent heat of vapourization for O₃ = 75.59 cal/g [5], and the heat capacity of O₃ gas = 0.195 cal/gK [6]; and assumed that the grenade is a sphere of radius 5cm, it starts at 90K, the man weighs 100kg, his heat capacity is the same as water, his body temperature starts at 311K, to be considered "turned in glass" his whole body must be equal to or less than 273K, the grenade only cools the man and thermal equillibrium is reached instantly, and no exothermic oxidation reactions occur that would heat the system. 91542 calories are required to bring the grenade to 273K, and that's enought to cool 100000g by 0.91542K for a final temperature of about 37C. So, no this isn't possible just by flash freezing. If only his surface must be cooled, than you'd get a bit further with the flash freezing argument, but then you'd want to start correcting for the other generous assumptions made above. 203.27.72.5 (talk) 01:03, 20 August 2012 (UTC)[reply]

Agreed, vitrification is possible, but not with a missile any man could heft in normal conditions. Better to use a hose. μηδείς (talk) 03:38, 20 August 2012 (UTC)[reply]

smell vs taste

Why does tea smell wonderful when dry but taste bitter once made? (Please don't digress into overbrewing tea etc - I'm talking about the correctly made stuff.) Ta, 184.147.128.34 (talk) 17:51, 19 August 2012 (UTC)[reply]

Tannins. --jpgordon^{::==( o )} 18:14, 19 August 2012 (UTC)[reply]

Coffee is the same way. And new tires smell good too, but I don't recommend eating them. ←Baseball Bugs ^{What's up, Doc?} carrots→ 18:49, 19 August 2012 (UTC)[reply]

Tannin#Drinks with tannins seems to disagree with you as far as coffee having much for tannins. BigNate37_(T) 19:57, 19 August 2012 (UTC)[reply]

So, tannins are bitter, but don't have a particularly bitter smell. I've found that herbal teas (tisanes) often contain fewer tannins, if any, so aren't as bitter, since they don't contain any real tea. This is good, since it allows me to drink them without sweeteners, either real or artificial. StuRat (talk) 20:21, 19 August 2012 (UTC)[reply]

Of course it's all completely subjective, I actually think the taste of straight tea is also wonderful, just like the smell. I don't like tea with sugar or milk added to it and I like it brewed quite strong. But I do occasionally enjoy a tea with lemon and honey, served hot on a cold day or cold on a hot day. Vespine (talk) 23:06, 19 August 2012 (UTC)[reply]

To actually comment on the question, I think it's just because smell and taste receptors are actually quite different in some fundamental regards. Even though a lot of the time they seem to compliment each other, like a lot of fruit smells precisely how it expect it should taste, but in other cases, like salt for example doesn't really have much of a smell at all. You can basically tip a cup of salt into a pot of soup and it won't smell much different, but if you tasted it, it would be completely inedible. Vespine (talk) 23:14, 19 August 2012 (UTC)[reply]

When I was kid, my brother baked brownies once. I first noticed when I smelled a beautiful intense aroma of vanilla wafting through the house. I came out into the kitchen to find out what he was doing and I asked him how much vanilla essence he put into the mix. He said that the recepie called for a cup but we didn't have that much so he just emptied the 50mL bottle that we had into the bowl. I had a look, and it actually called for a teaspoon. Needless to say, the brownies were probably the most bitter thing I've ever tasted. The look on his face when he bit into one of his creations was priceless. 203.27.72.5 (talk) 23:23, 19 August 2012 (UTC)[reply]

You need to distinguish between taste, flavour, and aroma. When it comes to taste, we're talking about what your tongue detects, which isn't as basic as once thought (and doesn't involve that stupid tongue map), but is a lot more basic than smell. The main tastes are salty, sweet, bitter, sour, umami, with probably a few more that we don't really discuss much. These are tastes, not smells, so you detect them with your tongue: you might guess based on past experience that something that smells of strawberries will taste sweet, but if you're drinking a fruit tea your tastebuds will be disappointed. You cannot smell how something will taste, beyond extrapolating from past experience. Bitter is a taste, not a smell.

When we actually eat something, we experience a flavour which is a combination of the taste and the smell/aroma. Because your nose and mouth are connected, you can smell the thing you are eating. When you drink a cup of tea, you taste the bitterness of the tannins which really shouldn't be much, if you're brewing the particular tea for the time it should be brewed for as well as experiencing the complicated combination of aromas from the tea that you could smell without drinking it: this combines to give a characteristic flavour.

When we add artificial flavourings to food, these are aromatic compounds that make the food smell nice, affecting the flavour: altering the taste involves much more basic things, like altering the amount of salt, sugar, fat, caffeine, etc, which also alters the flavour. A strawberry has a different flavour to a peach because it smells like a strawberry: it has different aromatics. Both fruits taste sweet. 86.157.148.121 (talk) 13:07, 20 August 2012 (UTC)[reply]

Just a silly story just in relation to how smell and taste are different, you mention fruit tea and I know exactly what you are talking about, but even more extreme example was this liquid fruit soap that I bought once which smelled exactly like delicious apples! But indeed tasted just like soap. Vespine (talk) 00:28, 21 August 2012 (UTC)[reply]

Thank you 86.157.148.121 that was a great answer; I get it. 184.147.128.34 (talk) 09:16, 21 August 2012 (UTC)[reply]