studiot

Well that's exactly it. Artistic techniques (which produce Art) can be used, very effectively for communication of ideas. An example of where you would not have swansont's cut and dried Newton's law would be found in my paper concerning cracks in concrete. The presentation version of the paper contains several animations of crack patterns and their development in concrete. These greatly aid communication to the subject matter. Cracks vary, every crack pattern is different, but characteristics of similarity can be extracted, from which much useful information can deduced. In case you are wondering; Cracks in concrete duh! So What? Well unwanted cracking has cost hundreds of millions or more worldwide so the subject is big, big business, as well as interesting science.

Looking quickly at the material in post#1 I can't see any reason for awarding a -1 vote, so I am going to add a +1. The material itself does seem to refer to highly suspect mystic nonsense. I had to look up 'sonic-geometry' on google to get an inkling of what you are discussing. I watched the beginning of a half hour video of challengeable claims about geometry and stopped at the point where they claim 'The basis of all angle measure is 60.' Which is, of course, just not true. I suggest you look for material that has been much better studied, developed and more rigorously tested such as group and symmetry theory. This is utterly fundamental to our current best way of viewing and understanding Nature. And the best part is that, compared to most branches of mathematics, it is really simple. That is it's great stength and beauty. You do not have to try to get your head round something like relativity to understand it.

I deliberately used the word particles, not molecules. This brings up several interesting related points. Gravity does indeed cause aggregation of particles. Some cosmology theory relating to heavenly body formation and growth relates to this. However gravity is not as strong as molecular bonds. If this were not so then any object you picked up would fall apart under gravity. Conversely gravity is not strong enough to put them back together either. John Cuthber has pointed out that breaking many 'solids' (particularly crystalline ones) involves breaking molecular bonds. Many materials are mixtures and indeed the lesser-than-molecular forces that hold these together will allow separation under gravity. So I said particles because particles includes molecules and larger (though still minute) material aggregates. Which comes back to surface energy and fracture reconstruction. Yes you would need to extract energy to perform reconstruction. You would also need a mechanism to perform this. Remember also that, as swansont has already pointed out, oxides form rapidly on fracture surfaces. This is because if you do break a molecular bond in fracturing, then you have an unattached bond end seeking a home to balance the valency of the molecule and the ever present oxygen in the air happily supplies this.

That's good since the same situation applies to all matter in all states. The particles at the surface are not in equilibrium. The interior particles are. And yes an ocean has less energy than all the water spread out into droplets and the larger the free surface of anything the more energy it has, to maintain this disequilibrium. This is why droplets try to pull themselsves into balls - to minimise the surface area to volume ratio. Breaking and reforming are the converses of this manifestation of surface energy.

Do you understand how surface tension arises in liquids? If not I will draw a helpful diagram.

Even though the pieces appear, to the naked eye, the same after the break as before, they are not. When you break something into parts you create new surfaces. This takes energy input. Every part with a new surface has more energy than before, by the amount of this additional surface energy. The theory of fracture mechanics is based on this fact and evaluating this energy and its effects.

You need to realise that the calorific value is the total amount of heat obtainable by complete burning of the fuel until no burnable material is left. Thus there can be no product B that has any further calorific value. For instance the complete combustion of carbon results in carbon dioxide, which is incombustible. No further heat can be obtained from the dioxide. Partial combustion results in carbon monoxide, which can be burned further to carbon diode with the release of more heat. The total heat avilable is the same whether you do the combustion completely in one stage to dioxide or in two, first to monoxide and then to dioxide. This heat is the calorific value of carbon. Enthalpy is a different concept entirely, although it is often measured in a calorimeter, and takes into account other forms of energy change such as mechanical work. So what would you like to do now?

Well yes, for some reason that is all they seem to mention in schools. But really there has been tremendous development in chemistry during the second part of the twentieth century. I am reading a really good book at the moment Introduction to Surface Chemistry and Catalysis by Gabor A Somorjai (Wiley ISBN 0-471-03192-5) You should find this much easier to find than the last British book I recommended. Fantastic developement of surface chemistry since I studied at university in the 1960s. Structural Chemistry and Materials Science in general have both also proceeded at a pace in the second half of the twencen. Analytical chemistry has changed beyong all recognition from the early twencen to the latter part. Chromatography, spectroscopy, Xray microscopy, .... In the first part of the twencen organic chemistry looked like Mrs Beatons cookery book. Pies, Cakes, Puddings, Roasts etc Alkanes, Alkenes, Alkyn, olefines and other homologous series. This has all been replaced by functional groups and reactions - nucleophilic, electrophilic, substitution etc. This has all lead to Plastics, carbon fibres, nanotubes, fancy drugs, kevlar, modern agrichemicals, a revolution in paint chemistry, a revolution in cement chemistry, chemical waste processing, the discovery of long chain compounds using other atoms than carbon ....... the list goes on What's your brew?

Did you see this bit from the Boeing link About carbonfibre plus aluminium?

I can only suggest you re read your own thread title, where you asked "How to evaluate.....?" I am trying to tell you how to do just that in accordance with currently accepted definitions; neither you nor I have the power to change these at whim. Your equation is not an equation at since since the left hand side does not, by your own words, equal the right hand side. I live 150 miles from imatfaal. What do you think he would say if I said to him? "just pop over on your bike, it's only 10 miles - the other 140 can be discounted because they are boring and of no consequence."

Beware your bike is not lost in the Southern Ocean. (Poor taste I know, but I couldn't resist it) http://www.boeing.com/commercial/aeromagazine/aero_07/corrosn.html http://www.bikeforums.net/road-cycling/222592-carbon-aluminum-don-t-mix.html http://www.boatdesign.net/forums/materials/corrosion-between-al-carbon-2609.html http://www.rcgroups.com/forums/showthread.php?t=1652668 Road salt?

What did you not understand about what I posted. I did not say anything about changing the calorific value. imatfaal has already told you (as I have) that your problem is in the belief that burning 1kg of A produces only 1kg of B. That is not true for any substance in the known universe. Perhaps you do not understand that the 1kg of A refers only to the fuel. But the number of kg of B refers to everything that results from the combustion. So in my example A = 1 kg hydrogen B = 9kg of water The burning chemical reaction adds 8 kg of oxygen to the 1kg of hydrogen. But the calorific value is still calculated on the base of the mass of fuel (hydorgen only. I will be out on site for the rest of the afternoon, so perhaps one of the chemists here might make this more plain for you.

I'm sorry I don't understand you last remark. As an example 1kg of hydrogen burns with 8kg of oxygen to yield 9kg of water plus C joules of energy. The definition of calorific value (of hydrogen) is C joules per kg. There is no escaping this.

2.7777 x 3600 kwhrs = 10 thousand kJ = 10 MJ I am taking the comma as a continental decimal point. But, dr2much The products of combustion always weigh more than the weight of the fuel.

Read this part of my post and what follows again. The product B is composed of some mass from A plus the mass of the oxygen needed for combustion. For example burning 1kg of A produces between 5 and 15 kg of B depending upon the fuel.

I can't comment on the numbers in your subsequent example as there is not enough information. But the above quote is a sound basis to start. Both A and as Sensei pointed out atmospheric oxygen, have chemical bonds which are broken and release energy. The combustion results in B, and some of this released energy goes into forming the new bonds in B. Some of this energy may go into a change of state of, if for instance the fuel is coal or oil and the result is carbon dioxide and water. What is left over appears as heat and is converted to electricity, ignoring the inefficiency of the generation process (this is far from negligable). Further, what Sensei was probably trying to say is that 1kg of A plus some kg of oxygen makes more than 1kg of B. That is you need to take the mass of oxygen into account in your reckoning. The relative masses of hydrogen, carbon and oxygen are 1, 12 and 16 So for every 1kg of hydrogen completely burned 8kg of oxygen are required. For every 1k of carbon, 16*2/12 = 2.7 kg of oxygen are required. So you see the oxygen soon mounts up to a significant portion of the mass of B.

I am not sure what you think is modulating what to create beats. Nothing actually travels instantaneously, I believe I said it was sensibly so. Consider the distance from A to B is about 0.3m An elastic signal travelling in the steel travels at 6000 m/s An elastic signal travelling in the air travels at 340 m/s The bell in struck with a hammer (at A), not continuously excited. So at the moment of imapct at A the part of the sidewall at A deflects. This causes a local pressuse disturbance around A. At this instant B is undeflected as is the local air around it. From the diagram you can see that the disturbance in the steel has to travel about the same distance as the disturbance in the air (=0.3m) So in the time for the disturbance in the steel to reach B and cause an elastic deflection there the distrubance in the air around A travels 0.3 x (340/6000) = 0.017m so it has not reached B, by a large margin This is why I am suggesting that the elastic disturbances at A and B in the air can be considered contemporaneous. This is an additional step to the usual analysis you will find for a piston in air because the piston is being driven at all points simultaneously. What the above shows is that the impact at one point on my bell has the same effect as if the whole bell was excited at once. The usual analyis a la a piston can then be applied and it become a question of geometry. That is the size of the radiator relative to the size of the wave. When the radiator size is comparable to (or much bigger than) the wave size, an off axis listener can receive waves from different parts of the radiator at different times because she is at different distances from different parts of the radiator. These waves can interfere and cause directionality as I previously noted. This effect is most marked at right angles to the main axis. But remember that the size of my bell is smaller than the wavelength of the signal I am observing. I can draw some helpful diagrams if you are interested.

Oh dear, rktpro, yes the equation is wrong. That blooming velocity, c keeps turning up in the wrong place. Unfortunately I don't speak LaTex so I write all my equations out in MathType and then paste them into the text. This time I wrote it in Word and pasted them in there and finally pasted it again into SF. Mea culpa. The correct equation is [math]Y = B\sin \omega \left( {\frac{x}{c}} \right)\cos \omega t[/math] The original two equations should have x/c as well. This does not alter the points made, however since the point is that the original waves contain additions (or subtractions) within the angle, of a time and distance function, reduced to common units. The composite of the two waves into a single standing wave results in the separation of time and distance into the product of a time function and a distance function.

Good morning function. All sound sources much smaller than the wavelength can be considered as an equivalent sphere. There is indeed much theory using equivalent spheres as it is important to the theory of both binaural hearing and stereophonic sound reproduction. As frequency rises the wavelength gets shorter and the sound emitted gets more directional. The higher frequencies contain all the directional information. A spherical wave contains no directional information. This is why it does not matter where you place a bass loudpeaker in a room and indeed you only really need one bass speaker. It is also why animals, such as bats and dolphins, use high frequency for location. For interest here is a short table of frequency and wavelength 50 Hz 6.90 metres 500 Hz 0.69 metres 1000 Hz 0.34 metres 5000 Hz 0.07 metres 15000 Hz 0.02 metres You can see that at 500Hz it need a pretty sizeable source to be bigger than the wave.

You might like to look at Ehrenfest's Paradox. http://en.wikipedia.org/wiki/Ehrenfest_paradox

What exactly am I supposed to make of this? What is it a reply to, did you not read my post? Do you think 6000 is a few thousand? I would say a few thousand is more than 1000 but not more than 3000. This is the figure you will find published as I noted. I further note your Wikipedia link is self admittedly half finished and does not publish any figures.

Setting aside all the personal remarks I will take one statement and ask you for detailed backup. The figures I have quoted were published by the Applied Research Laboratory of Pennsylvania State University in Acoustics Today. If you disagree with these then I look forward to your detailed alternative figures along with your substantiation of them. I am still looking forward to your detailed mathematical analysis requested in my post#16.

You have alluded several times to my educational background. It seems that not only did we have different professors of mechanics at university; we also had different professors of etiquette. I have not once disputed that the bell has flexural modes of vibration, or that the one we are both concentrating on is flexural. In fact I described such vibrations very early on in posts# 4, 8 and 10 and posted links to many pictures of the same. I did not explicitly state the mode of vibration, just described it as complex, which it is, although this information was available in the links. It was a good addition to the thread for you to state this explicitly as flexural or bending in your post# 15. Flexural vibrations are not sound waves. Sound waves are pressure waves that travel at the speed of sound in the medium. Flexural waves have a different speed and equation. However we are discussing sound in the air, not in the bell. OK let us examine the mechanics of the situation. I stated in post# 10 that the bell launches a spherical wave into the surrounding air. This is in agreement with my measurements around my test bell. That is the readings possess spherical symmetry. There are no measurable nodes or antinodes and the wave is not, as you stated in post# 21 composed of standing waves. There is no measurable interference. This implies that the bell should be considered as a single point source, emitting a single expanding spherical wave. To see how this can arise and why the relative speeds of sound are important look at diagram (1) Points A and B are on opposite sides of a bell, which is immersed in undisturbed air. The constant that has the dimensions of velocity in the differential equation describing flexural vibrations of plates and bells has a value of a few thousand m/s at the frequencies observed. The speed of sound has a value approaching six thousand m/s, as already noted. The theory of elasticity requires that any elastic disturbance, imparted at A, is transmitted as rapidly as possible through the metal of the bell to B. It further requires that the maximum speed is the speed of sound. So any disturbance imparted at A will create a disturbance locally in the adjacent air. The disturbance through the metal of the bell will reach B some 10 to 20 times faster than the disturbance around the outside of the bell since the airborne disturbance travels at 340m/s, since both paths are essentially the same length. When the disturbance in the metal of the bell reaches B, the elastic response at B generates a disturbance in its own right in the air local to B. The point about this is that the response at B is effectively instantaneous so B can be considered to be launching its part of the wave at the same moment as A. Another way of putting this is to say that the wave at B is in phase with the wave at A or that time zero is the same for both points or that the wave centres from A and B are coincident. This is important because it is our justification for the statement that the bell is a single point source generating an expanding spherical wave centred somewhere in the middle of the bell. It is also in complete accordance with conventional wisdom that sound sources smaller than the wavelength of the sound generate spherical waves based on a single point source. So we have an expanding sound field generated by the bell acting as a single point vibrating source, that looks like diagram (2). There is no interference since a single wave cannot directly interfere with itself. Which brings us nicely to the issue of interference. The principle of superposition states that the resultant motion is the sum of the individual motions acting separately. So for any two waves Ytotal = y1 + y2 Now a spherical acoustic wave has both real and imaginary parts in its equation of motion. But I am going to take a plane wave since it has only real parts for simplicity in the next analysis. [math]{y_1} = {A_1}{\omega _1}\left( {x - ct} \right)[/math] [math]{y_2} = {A_1}{\omega _2}\left( {x - ct} \right)[/math] In general since omega1 is not equal to omega 2 the sum of these two travelling waves yields a modulated travelling wave. In general the sum may not have periodic zeros. However by suitable choice of constants and making omega1 equal to omega2 we find after a bit of trigonometric reduction on the sum [math]Y = {y_1} + {y_2} = B\sin \omega x\cos \omega ct[/math] This is the equation of a standing wave since the first (sine) term has periodic zeros distanced in space between points in space specified by omega times x, regardless of the time t. The point is that the wave (a cosine) is multiplied by a sine wave, which has periodic zeros. Thus this equation is the equation of a standing wave and is how nodes arise. Now you have stated categorically that standing waves interfere. So please show mathematically how the superposition (sum) of two such waves can interfere to form zeros or nodes. You have also mentioned that the harmonics of the vibration die away more quickly than the fundamental. Yet you claim energy analysis is irrelevant at best and misleading at worst. The energy in the harmonics of an oscillator increases with harmonic number. The higher the harmonic more energetic it is. Thus as energy is lost to the vibrating system through dispersion and/or dissipation, the energy to power the highest harmonic is lost first. And so on down the list. This is commonly observed in vibrating systems. OK so we have generated an interference free expanding spherical wave in the room, but we are here to explain the why the sound level is lower inside the bell. What is different? The only thing that is different is that the air inside the bell is significantly more confined that outside. Here are a few calculations to show how this makes a difference. The room volume is V0 = 3x2x10 = 60 cubic metres The bell volume is [math]{V_2} = \pi \left( {{{0.2}^2}} \right)\left( {0.15} \right) = .02[/math]cubic metres And the bell surface area is 0.2 square metres Say the average movement is 0.01 mm = 2x10-5 metres Then the volume change on compression is –2x10-6 cubic metres (note negative for compression) Now the volume and pressure changes follow the adiabatic gamma law with gamma = 1.4 So For the room [math]{P_0}V_0^{1.4} = {P_1}V_1^{1.4}[/math] and for the inside of the bell [math]{P_0}V_2^{1.4} = {P_3}V_3^{1.4}[/math] Since the original pressure P0 is the same in both equations solve each for P0 and equate. [math]\frac{{{P_1}}}{{{P_3}}} = \frac{{V_3^{1.4}V_0^{1.4}}}{{V_1^{1.4}V_2^{1.4}}}[/math] Now if we observe that V0 = 60 cubic metres and V1 = (60 - .000002) cubic metres we can approximate V0/V1 = 1 Thus [math]\frac{{{P_1}}}{{{P_3}}} = {\left( {\frac{{20000}}{{20002}}} \right)^{1.4}} = 1.00014[/math] Which very clearly states that the pressure generated outside the bell (P1) is greater than the pressure inside (P3) and of the correct order since sound pressure levels are measured in parts per million relative to normal absolute (1bar). But we should further note that these are absolute pressures in order to use the adiabtic gas laws. The sound pressure level (spl) is actually a pressure difference between the quiescent pressure and the wave pressure. That is spl outside = (P1 - P0) and inside = (P3- P0) We could calculate these by using a standard value for P0 eg (1bar) and I will leave that as an exercise. So is detailed discussion of the modes of vibration of the bell or standing wave interference really necessary to answer the question?

73% of 50 is exactly 36.5. Where you count the maybes as yes or no where does the half a response come from? Hence the comments in the other posts.

Was the moisturising lotion London Pride or Fullers Winter? Both these also come in half pints for those who get half cut and make half _ssed statements.