studiot

All of Strange's comments are possibilities, to these I would add Think about a cheap transistor radio v a hi fi system. It is a well known acoustic phenomenon that a cheap low power radio can sound as loud as an expensive hi fi simply because the ear is particularly sensitive to distortion. The radio sound is much more distorted than the hi fi and sounds louder, even though the power is lower.

I thought ajb did a pretty good rundown of theoretical. I am an applied man myself, but isn't non linear stuff pretty theoretical?

Was your diagram something like this? What is the definition of cente of gravity in relation to the weight or mass of an object? I am not going to do all the work for you so I have left some blanks on the diagram for you to fill in. Ask if you need more help, but you should be able to post an equation balancing the moments of the three forces and leading to a quadratic equation in d.

Have you drawn a diagram? Where is the centre of gravity of the chunk of metre stick to the left of the fulcrum?

You have made it to simple, trying to combine too much. You have three moments to balance. Each has a different lever arm. Here I will start you off. I strongly suggest you draw a diagram. Working with zero art the left hand end Let d cm be the distance from the zero mark to the fulcrum. So the distance from the fulcrum to the 100cm mark at the right hand end is (100-d) cm The mass of the ruler is 0.24kg ie 0.024 kg per centimetre. So the mass on the left hand side producing counterclockwise moments is 0.024d kg The counterclockwise moment is therefore (0.024gd) x(d/2) N-cm. Can you now complete by working out the right hand side clockwise moments properly?

Thank you , Larry for your response. I wonder why you did not read it closely? Here is an extract from the generally lucid Wiki article: Perhaps you would like to stop and ponder how or what it means for a passive solid body to 'exert a force on the air' The look again at my post#93 (it was a short one). Perhaps this is why momentum is considered rather than force in fluid texts? I have not said that there is no downwash. Quite the opposite, although I did not call it that. The downwash is the 'down' part of the circulation (surprise surprise)! Pressure differences above and below the obstruction, in front of it and bhind it as well? Of course there are. I have never said any different. But what pressure are you discussing, Dynamic pressure or static pressure? And are you aware of the huge variation of these along the section of the obstruction? There are some much better diagrams in my link in post#91 than the ones in the Wiki article. As regards the Wiki article application of Bernoulli's theorem, they are discussing horizontal flow in an infinite atmosphere. This is not the correct statement of the theorem or application of it to post#83. As a matter of interest the correct application of Bernoulli's theorem in the Wiki case should include the circulation, a point often missed in explanations including Wiki itself. This Wiki case is actually much more difficult that that of post#83. You need to fully understand Bernoulli's theorem to apply it correctly. What did you make of the questions I posed in post#98, in particular this one? A while ago we had a very long thread by someone who insisted that a particular 'explanation' of airfoil action was due to certain physical laws only and the others were wrong or didn't apply. Of course the truth is that they all always apply so if one apparantly contradicts another there is something amiss with the analysis. Sensible persons use the easiest set of laws to obtain the answer, not the most loved or the most difficult. This is the situation in many walks of mechanics.

Sorry I wouldn't have t' foggiest. Us grunts just get on with it. Rules and regs we leave to the lawyers who like that sort of thing.

Hello Daniel did you understand that stuff about vectors, dot and cross products? I ask this because there are two units that are the product of force (Newtons) and distance (metres) Force and distance are always vectors in this context so when we multiply them two products can be formed. Most folks use the Newton-metre to refer to the vector product or moment. This is still a vector. Distance in this case is taken at right angles to the line of the force. So far as I know we haven't graced anyone with this unit and the unit of moment is only the Newton-metre. Perhaps we should call them the Archimedes? The alternative multiplication is the cross product and the result is a scalar we call work. Work has the same dimensions and unit as energy - the Joule. People seem to use both lower and upper case j's for this unit indiscriminately. Also the Newton metre is not often used for this unit.

As regards to the 'experiment' in post#83 Thermal transfer due to friction and turbulence are simply red herrings. In this experiment the OP has chosen to soapbox and simply demonstrated his misunderstanding of energy conservation in general and Bernouilli's theorem in particular. This experiment can be addressed by a correct application of the simple form of Bernoulli's theorem, containing three terms, such as would be taught in the first semester of an engineering fluids course. I very much doubt that turbulence is involved. Does anyone know if Bernoulli's theorem can be applied to turbulent motion? Bernoulli's theorem is an energy balance for any point in the fluid and it states that: The sum of all the energies of the fluid = a constant minus any work done by the fluid In this case there are three energy terms The static head, the dynamic head and the gravitational head. At the outset the first two are zero and no work is being done. This establishes the value of the constant. When the fluid is initially allowed to flow no work is done and all the gravitational energy head energy is exchanged for potential and dynamic head. When the obstruction is released if it rises at all, and it may not, some work is done and so the constant in the Bernoulli equation is reduced by this amount. Now, oh wizards of fluids 101, 1) Is it possible to solve this system and distribute the changes in the three fluid energy terms without further information other than fluid physical properties? 2) I noted that the obstruction may not rise at all, even when it is free to do so. What controls this? As regards to thermal transfer. Consider an aircraft that takes off from the ground at 20oC and climbs to 30,000 feet. Why does the pilot worry about wing icing if the object is heated by friction? I look forward to Zet's correct Bernoulli analysis.

One very important question to answer, if you are considering an energy analysis. It is easy to state that the 'energy comes from the fluid', but which bit of fluid? The fluid above, below or to the side of the airfoil? The fluid that has yet to arrive at the leading edge of the airfoil? The fluid that has already passed the airfoil? The fluid that is just passing the airfoil? All the fluid (in the atmosphere)? Some combination of these? It is easy to state that the energy loss causes the fluid to slow down. But wait, if the fluid that has passed the airfoil is now moving more slowly than the new fluid arriving, then fluid is building up somewhere in the system. But the fluid has been declared incompressible. Continuity (Laplaces equation) is violated. And if the energy is being transferred from the moving fluid to the airfoil, what proof is offered that the agent driving the fluid doesn't simple drive harder (ie add more energy to the fluid) to replace the transferred energy? Any who really want to understand the energy interactions of flight need to be able to consider and answer these questions and others. Controlled flight is a very complicated system and attempts to simplify usually remove some essential link in the chain, thereby breaking the whole chain. I think this is the problem with Zet's attempt at simplification.

Airfoil action is not to do with gravity. Discussion of bouyancy only clouds the issue airfopil lift and bouancy are very different and not mutually exclusive. Air moves up as well as down, hence the term circulation. It is the circulation that cause the lift and the interaction of the obstruction (airfoil) and the moving fluid that causes the circulation. Circulation is theoretically possible in a frictionless fluid, but uncontrollable. It is friction that allows us to control the circulation, thereby achieving consistent and sustained lift. I have been trying to find a simple explanation wihtout higher mathematics. Here are some nice animations. http://www.diam.unige.it/~irro/profilo4_e.html

arc, your comments would be fine except for one thing. Yes I said earlier that a glider will eventually return to Earth (thereby loosing its potential energy) and that is true here as well. However Zet's (correct) point was that this energy must have come from the fluid and therefore the fluid must have lost some energy. The point is that the story does not stop there, since some agent must have input the energy into the fluid in the first place or it would not be moving. Further continued lift requires a continued supply of moving fluid and therefore continued external energy input.

Pun intended.

Thank you endy, for the continued support and explanations, which are very helpful. That (helpful) is the difference between humans and programs. IMHO humans invented ergometrics as well as programming, but forgot to teach the former to the latter.

Let f(x) = a [math]\forall [/math]x Then f(a) = a thus f(f(x)) = a [math]\forall [/math]x Let g(x) = a for x=a and be undefined otherwise. then g(x) = f(f)x)). But this still avoids the two points I asked you directly about. Are you refusing to answer them? Nice one imatfaal +1

But you didn't say so. Which is why I called your statement incomplete. Fo one function to exactly match another they must share the same domain and codomain. You haven't specified this either. How will you make that work for f(f(X)) ?

I asked you (politely) to explain in more detail you meant by your "requirement" and gave an example of why I consider your statement f(f(x)) = g(x) incomplete. You have made no attempt to do so that I can see. So I offered an example for which there are an infinite number of values of x for which f(f(x)) = g(x) Do you deny that this is so and if you deny it can you prove it mathematically?

Since I am addressing your points, either you address my points as well or we will need to discontinue this conversation as one sided. I said For an infinite number of x, sin(x) = tan(x). It is also true that for all these x, sinx = tan(tan(x)) So your claim is nonsense.

You did indeed write that, but did not explain why, so I took it as a typo. Sorry. It seems a bit pointless since a simple substitution reduced f(f(x)) to q(x) and we have the situation I have described (in post 51) and you need to elaborate on.

That didn't answer (=avoided) the question of what you mean by the statement f(x) = g(x) So how can anyone agree or disagree if they don't know exactly what you mean? (hint the above equality statement (equation) is incomplete as it stands) I will be helpful this once, although the onus is on you since you are the one trying to prove something. The functions g(x) = sin (x) and f(x) = tan(x) are equal for an infinite number of values of x Unfortunately they are also not equal for an infinite number of values of x They are also very nearly equal for an infinite number of values of x They are also as different as it is possible to be ( that is f(x) - g(x) = [math]\infty [/math] for an infinite number of values of x If h(x) = x then h(x) = f(x) = g(x) for one value only of x. So what exactly do you mean? And you have still avoided my example points.

I have already expressed my opinion as to what impedes agreement. We have understood you , but you are making no effort to understand us. You would have to be more specific as to what you mean by f(x) and g(x) and what you mean by the equality of two functions.

No it does not. But that is totally beside the points I made, which you are still avoiding.

Writing stuff down is good +1 Rearranging it later is even better. There are lots of educational studies that show comprehension and retention is better if we write stuff down than if we just read or listen to it. I once had a teacher who used to keep saying "The more times you write something down the more danger there is that you will remember it" As to what to write down, the only data constants I would remember is that the acceleration due to gravity, g is approximatedly 10 and that pi is approximately the square root of 10. In the light of your other thread about exponents, it is not clear at what level you are entering science education however you will find that greek letters are much used so if you can become 'comfortable and familiar' with these it will be a great help when you encounter expressions such as [math]{\tau _{\max }} = \frac{1}{2}\sqrt {\left[ {{{\left( {{\sigma _x} - {\sigma _y}} \right)}^2} + 4\tau _{xy}^2} \right]} [/math] and [math]\left( {\begin{array}{*{20}{c}}{{\sigma _{11}}} & {{\sigma _{12}}} & {{\sigma _{13}}} \\{{\sigma _{21}}} & {{\sigma _{22}}} & {{\sigma _{23}}} \\{{\sigma _{31}}} & {{\sigma _{32}}} & {{\sigma _{33}}} \\\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{{\sigma _x}} & {{\tau _{xy}}} & {{\tau _{xz}}} \\{{\tau _{yx}}} & {{\sigma _y}} & {{\tau _{yz}}} \\{{\tau _{zx}}} & {{\tau _{zy}}} & {{\sigma _z}} \\\end{array}} \right)[/math] Don't worry what the letters stand for, just don't be put off by their use. The matrix brings us to Mathematics. You can't do much Physics without it, so you will quite quickly need to move through a lot of maths. Examples are Complex numbers Trigonometry Matrices (as above) Simultaneous equations If you have done any calculus Taylor and Maclaurin series Agin on first encounter it is enough to be able to recognise 'Oh yes, that is a matrix. That is a Taylor series, That is the cosine of the included angle' and so on. Go well in your future studies.

I have no idea what you mean. The original matrix was generated in MathType and pasted in between math tags. The <br> and anything else was added by the system without being asked or told.

Well rearranging what I wrote doesn't seem to work so what exact amendmends do I need to the code please? [math]{\begin{array}{*{20}{c}} {{a_{11}}} & {{a_{12}}} & {{a_{13}}} \\ {{a_{21}}} & {{a_{22}}} & {{a_{23}}} \\{{a_{31}}} & {{a_{32}}} & {{a_{33}}} \\ \end{array}}[/math] [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] Finally got it to work. Thanks a bunch endy +1 What a cack handed piece of duff programming. A machine should do what it is told period.