studiot

Well it partly depends what you mean by important and also what you include in classical mechanics. Are you thinking in purely theoretical terms or do applications count? There is much work yet to be undertaken int the field of non linear mechanics. You have listed several application areas which are the most heavily studied. but how about The mechanics of granular assemblies The mechanics of fracture and fatigue The mechanics of fibre bundles and other composites High and low termperature mechanics Near vacuum statistical mechanics the mechanics of non linear wavelike motion

How do I get rid of the unwanted <br> whatevers in this matrix code please? [math]\left( {\begin{array}{*{20}{c}} {{a_{11}}} & {{a_{12}}} & {{a_{13}}} \\ {{a_{21}}} & {{a_{22}}} & {{a_{23}}} \\ {{a_{31}}} & {{a_{32}}} & {{a_{33}}} \\ \end{array}} \right)[/math] I have left the brackets of the math / math tags so the code will show up. math\left( {\begin{array}{*{20}{c}} {{a_{11}}} & {{a_{12}}} & {{a_{13}}} \\ {{a_{21}}} & {{a_{22}}} & {{a_{23}}} \\ {{a_{31}}} & {{a_{32}}} & {{a_{33}}} \\ \end{array}} \right)/math If I paste this code into Codecogs the matrix is correctly displayed.

That is close enough. +1

If you place an onject at the base of a wedge and push it horizontally onto the wedge what happens? So you confirm that it is a waste of time talking to you. So this is the last time I will tell you: If some external agent does not put enough energy into the system for flight you cannot take it out.

What will happen to any liquid exposed to an extensice vacuum at its free surface as you have drawn it? It will evaporate very rapidly. If you weren't so determined to proove yourself absolutely right and everyone else absolutely wrong you might actually learn something. D'Alemberts Paradox, (some call it D'Alembert's Theorem) refers to ideal fluids. That is incompressible fluids without friction. Briefly put, it states that ideal flow past any object creates zero drag in the direction of flow. So you see this is highly relevent to your simple airfoil model. The question in post#46 was a restatement of your original question and has been answered many times by several posters in this thread. Your question is basically "Where does the energy to lift the airfoil come from ?" The energy comes from whatever source drives either the airfoil through the fluid or the fluid past the airfoil. However it is really difficult to match a particular parcel of energy in the fluid with a particular amount of lift energy. I will try one more time by offering this diagram and very simple explanation. Edit to be completed. OK so how much energy to raise the airfoil unit distance ? This one is easy it's airfoil mass times 1metre times g But how do you relate that to the fluid? That's not so easy since each parcel of fluid is blowing past the airfoil and adding in a small part of this energy before it is replaced by another parcel. And how far into the fluid do you consider your airfoil drawing energy? Well one way is to calculate the lift force. To do this look at my diagram. I have drawn two streamlines BC and AD and two more (B'C' and A'D') exactly 1 metre alongside so they form a square box section stream tube. The streamlines are above and below a 1 metre section of airfoil. So everything in the third dimension is measured per metre. The fluid is considered ideal which means it is incompressible and inviscid. The fluid is approaching section ABB'A'A horizontally so its velocity, U equals Ux its horizontal velocity and its vertical velocity is zero at this section in front of the airfoil. Physical observation shows that the streamlines turn downwards as they pass the airfoil as shown. So when the fluid emerges from section DCC'D'D they have a horizontal and a vertical velocity. Now the volume rate of flow through each section = area cross section times the velocity and since both sections are vertical (sorry about my sketch) they have they admit zero volume flow vertically. Since the fluid is incompressible the volumetric flowrate through each section is the same. U1A1 = U2A2 But both sections have the same area so U1x = U2x This is the obstruction does not slow the fluid horizontally, in accordance with D'Alembert. What does happen is a vertical downward velocity Uy is imparted to the fluid. The mass flowrate equals the fluid density times the volumetric flowrate, and the momentum flowrate equals the mass flowrate times the horizontal or vertical velocity. At section1 there is zero vertical momentum in the fluid but at section 2 there is a downwards momentum flow of (pUxA2)Uy But a momentum flow is the rate of change of momentum and this is the definition of a force. The force exerted on the airfoil is the lift force and this is equal to in magnitude but opposite in direction to this force on the fluid. So the lift force = Fy = -(pUxA2)Uy This is the same formula obtained in Jukowski's theorem by complex integration, but obtained much more easily, if clumsily.

Good Morning, I hope the wedding went well. Back to geology; I'm sorry I find that explanation as clear as mud. Perhaps you mean former compression? Otherwise you need to account for the fact that the Atlantic floor is currently widening and cannot therefore be in compression, rather than tell us about the Himalayas whcih are not in your section.

In my first year in University, my roommate used to go to bed every night muttering the Chemical Engineer's mantra. He attributed his high success in the end of year exams to this being the only thing he could remember and he applied it remorselessly. Input = Output plus Accumulation Since then I have been conducting successful engineering balances for over forty years, force balances, momentum balances, energy balances, mass balances..... So please don't lecture me on how to conduct a balance. The trick with balances of any sort is to know what to apply it to. If you try to apply it to the whole system as you are, you will often miss out important contributors, as you have done. At least you have now understood what the balance applies to, although I explicitly stated what energy changes I was referring to. Congratulations on realising that the body must bob up and down for a period. However this will not be forever since you have neglected the atmosphere above the container. Both the surface of the liquid and the body itself will oscillate up and down, generating pressure pulses in the atmosphere above. This is a dissipative process that will eventually leak away all the kinetic energy in radiated wave motion in the atmosphere. Absolutely not. Horizontal motion is just that by definition. Before we discuss friction, have you heard of D'Alembert's Paradox?

Yes indeed not only I, but others as well have fully understood your points and responded showing that they do. Understanding has not been so fortunate the direction. I have shown you two examples where it is impossible to obtain any repeatable result or output from a mathematical equation. One of these was for reasons of Physics, One for reasons of Mathematics. You keep avoiding these for some reason of your own. Why is this?

And yet to shout at me for saying that friction is necessary. What is (almost) axiomatic is that any body that moves away from the commen centre of mass will gain gravitational potential energy, whatever the cause of of that movement. The gravitational field is a conservative field, which means that the difference of potential energy of any body between two positions depends only upon the positions and not on the path taken to reach them. That is axiomatic in potnetial theory. You seem so desperate to prove me ( and the rest of physics wrong) that you are not listening to what I said. There is a difference in the energy flows for the system, the rising body and the fluid (which incidentally I have never accepted neceessarily 'falls'. No fluid falls if you blow it horizontally past an airfoil.). You are confusing all these energy flows. Which is why I keep repeating that an energy analysis is the difficult way.

Whilst I am happy to discuss A or B type considerations with you, the real world, both of Nature and Mathematics admits of types C, D ..... If you cannot see these surely the logically correct course of action is to ask for further detail, rather than pretending they don't exist. Do you understand what 'no solution in closed form' means, for instance? If I am wasting my time addressing your points whilst you ignore mine please tell me as I have many other things to do. If you wish to continue the discussion I have shown you two additional situations, and there are yet more examples of non determinism and ways to cope with it. The next one would be limit state design philosophy. Do you know what this means?

The speed of light is a measured quantity. There is no theoretical derivation of the figure, although it is the ratio of other measured quantities. Here are the results of the first hundred years since we have been able to measure it.

If we take one point at a time we might get somewhere useful. Let us re-examine this in the light of some physics. Your block-in-a-bucket here is an example of an isolated system. This means that neither any mass nor any energy is allowed into or out of the system. So both mass and energy are conserved within the system. Now let us look at you contention that "There will be an increase in kinetic energy and an equal decrease in gravitational potential energy." So the block floats to the surface. Then what? Then it stops! So what now has more kinetic energy? Similarly some liquid falls. Again then what? Again it stops. So now we have a lower potential energy and zero kinetic energy in the system. So where did the energy go, if it was conserved? Keeping to the topic of potential energy you seem to be arguing at cross purposes with yourself here. No. It does not “impart” gravitational potential energy. It loses gravitational potential energy. Had your read and quoted the full of my text (was two lines too many) it would have made sense, in that it would have been clear that the word 'it' refers to the bouyancy force imparting (gravitational) potential energy to the floating object. You have emphatically denied this as shown above. You are happy to assert that falling objects loose gravitational potential energy, yet seem unable to accept the rising ones gain it.

You seem to only have considered two possibilities. Probability v Determinism Nature is more diverse than that, both swansont and I have mentioned situations other than probability where the mathematical equation does not have a nice tidy deterministic output. swansont mentioned the uncertainty principle which is a physics principle that the product of momentum and position or energy and time cannot be known exactly. As a result the more deterministic your calculation about one the less deterministic is your calculation about the other. This is nothing to do with mathematics, but to do with the physics of reality. I mentioned that many equations have no solution in closed form. That means that it is impossible to arrive at a perfect deterministic value for the output Y (given an input X) that the equation is calculating. Bessel's equation is a simple(?) example. Of course we can get as accurate as we wish (unlike with the uncertainty principle). This is inherent in the nature of the mathematics and nothing to do with the physical world that such equations are used to represent.

The physics of projectiles have nothing to to with the physics of airfoils. So please do not claim to be following the laws of physics as here Of course they must be correctly applied, which you have singularly failed to do. This must be your best (or worst?) pronouncement. No. If the one airfoil (as described in post #46) is able to move vertically but not horizontally then it is not friction that keeps it in place. It must be in some sort of track (a frictionless track it is now here so stipulated) that allows it to move vertically but keeps it in place horizontally. So are you really saying that is not physically restrained against the real physical drag force or are you just wasting everyone's time?

Thank you for your references, I will look at them shortly. Meanwhile please look at this discussion about the word 'random' http://www.scienceforums.net/topic/84215-chance-vs-probability/page-2 I extract my post#29 for convenience. Final point let me thank you on conducting an adult and professional discussion without the rancour too often seen here. You are certainly proving up on your original statement. and your points are not easy to answer.

I'm sorry if you did not understand this. It is basically the same point both strange and swansont have made. There is a deterministic equation that will yield a definite probability P(E) of some event E. Hence there is another definite probability of a different event = {1-(P(E)}. Now you started this thread with the proposition So, given P(E) or {1-P(E)} we are all asking what is the deterministic prediction of the outcome?

Not so by Godel's Theorem. You still haven't answered my second point in post#12

OK so geodesics. They have an interesting story. Wikipedia does a fair job of summarising facts. I have picked out some salient points and numbered them. http://en.wikipedia.org/wiki/Geodesic Unfortunately, Wiki gets the history wrong by nearly 2000 years in point (note 3). Geodesic is an Ancient Greek word ‘geodaisia’ which means ‘divides the Earth’ from two Ancient Greek Words geos- the Earth and daiesthai - to divide. To understand this you need to understand that in 2000+ years ago Science, Philosophy and Religion were inextricably linked. The Greeks of that time regarded ‘perfect’ shapes is proper and real and many cultures had a Royal Road that could only be traversed by the ruler-cum-deity. Since to them (some at least) a sphere was the perfect 3D shape, the Earth was a sphere. This road did not deviate to right or left from its path (kept straight on). So if one followed it one would eventually come back to one’s starting point and one’s path would ‘divide the Earth equally’. This accords with the notion in (note 2). They were, of course, talking about great circles on the globe. Moving on nearly 2000 years cartographers realised that the minor arc of a great circle on a sphere is the shortest distance between two points. Unfortunately they also realised that the Earth is not quite a perfect sphere, and the modern idea of a geodesic as the shortest distance on the real shape of the earth was born. So the mathematicians of the time got hold of the issue (which is where the Wiki history starts) and they realised that geodesics had other properties, besides shortest distance. (Note 1) shows that the proved that geodesics on a plane are straight lines and that lead to the mathematics of developable surfaces and ‘ruled’ lines. (Note 6) was the culmination of 100 years of post renaissance maths development, primarily in Europe where the geodesic as the shortest line between two points on a surface dominated. This period, to the mid 19th century, also saw the development of much of the apparatus of modern maths, in particular the idea of manifolds (note4) instead of surfaces. All these were variations on the cartographic idea of a geodesic, where the scale is even on all axes. This brings us to (note5) which introduces the next instalment of the story and provides the link to geodesics as minimisers (or maximisers) of other expressions besides distance.

No so. You are the one who said that X is the input and confirmed that X could be any number. The problem is that you cannot obtain an output although you can indeed write the equation 82 - 4 = 0 Which is what i actually said, written out as an equation. If it is of interest this is because the equals sign in this equation is differnt from the equals sign in the first one I presented. The equation, x + (-x) = 0 is an identity ; That is it holds for all values of X The equation X2 - 4 = 0 is an equality. Perhaps you have heard of and understand the difference? Secondly you have not answered my second example in post#12

So if I choose x = 8 as the input to this equation, what is the output ? x2 - 4 = 0 Your theory is also in difficulty with simple probability. I can state the probability of heads in a fair coin toss is 0.5. X = the probability of heads, output 0.5 Deterministic as you say. But what is the output of the equation X = the result of next toss of the coin? How is that deterministic?

I suppose the inputs to a mathematical equation must be other equations as a substitutions or numeric values. Mathematics as the language only deals with equations and numeric values, physics as an application of that language then adds concepts like types (e.g. cats and dogs). So, just to confirm, are you saying I can freely input any number to an equation and the output is always deterministic?

Billiards, that is a very good question +1

I noted today a question from the originator of a thread about a comment I had made. I had presented this, more than once, in the past so I tried a search on 'zero' first with my name attached and second without, to find my previous replies on the subject. Both searches returned a zero result and I consider this pun by the system most unfortunate.

You didn't (put numbers to it), I did. But variations in speed, acceleration, and distance must all be taken into account in an energy analysis (balance). You have not done this. This is part of what makes energy analyis not as simple as we might like. Here is another reason. Your comparison of bouyancy forces and airflow lift forces is flawed. No the pressure difference is not gravitational potential energy. If the (bouyancy) force leads to movement of the object then it will accelerate the object, imparting the kinetic energy of movement to that object. In addition, once the object has been moved upwards some distance (any distance however small) then it also imparts gravitational potential energy. All this is fine and dandy and leads to the following energy balance. As soon as a net bouyancy force exists (however small) it accelerates the object upwards (however slowly), thus imparting kinetic and potential energy to the object as noted above. However this does not happen with an airfoil lift. It is well known that an aircraft has to achieve a certain minimum speed in order to be able to take off at all. Translated into air motion past a stationary airfoil this means that if you free stand an airfoil on a support and blow air past it, there will be a minimum airstream speed, below which nothing will happen. An aside here to Larry jevens This is why we need friction. The airfoil would simply blow backwards off the support in the absence of friction as we brought the airspeed up above zero. So if we consider any airstream speed in the sub takeoff range, a lift force will be generated, but the airfoil will not rise. This is unlike the bouyancy force situation described above. No one doubts that the airstream imparts energy to an object in its path. But this energy is always initially purely kinetic. Any potential energy transferred comes later. The second question is "What is the form of the energy loss from the airstream?" Again this is complicated since the airstream may possess linear kinetic energy, rotational energy, static head (potential) energy, gravitational potential energy and thermal energy. This is why an energy balance is less than easy for flight calculations.

So are you saying I can input anything I like to an equation and get an output? Please note I mean valid mathematical objects, like numbers not cats, dogs etc, I am not being facetious.