studiot

I did ask for a sketch to confirm your flow. Bignose was also unsure. I think you mean what is called plane parallel shear flow given by the flow field. u = u(y,x,z,t) = (u(y,t), 0, 0) Note line 6 of post#2 should read The first equations tell us that p is a function of x and t only. Further since u is independent of x, [math]\nabla .u = 0[/math]

You should also take note that as waves progress to shallow(er) water the wave height increases and the profile departs markedly from sinusoidal, in particular the crest amplitude and trough depth are no longer equal or equal to half the sinusoidal amplitude, a. See the Reid and Bretschneider 'Breaking Index Curve'.

What do you mean by 'a straight line' and why is it important to you?

Have you tried trigonometric substitution. with x = tan2(t) and then use sec2(p)= (1+tan2(p))

Perhaps the correction to my last post will make more sense as an answer.

A picture is worth 1000 words. I think you mean [math]\frac{{\partial p}}{{\partial y}} = \frac{{\partial p}}{{\partial z}} = 0[/math] as conditions for [math]\frac{{\partial u}}{{\partial t}} = \frac{{ - 1}}{\rho }\frac{{\partial p}}{{\partial x}} + \nu \frac{{{\partial ^2}u}}{{\partial {y^2}}}[/math] The first equations tell us that p is a function of x and t only. The second (NS) equation tells us that [math]\frac{{\partial p}}{{\partial x}}[/math] is equal to the difference between two terms that are both independent of x. Thus [math]\frac{{\partial p}}{{\partial x}}[/math] must be a function of t alone. since we have steady state [math]\frac{{\partial p}}{{\partial x}}[/math] must therefore be zero. Am I reading you right? #Edit My apologies I see a serious clanger crept in during the copy/pasting. I said [math]\frac{{\partial p}}{{\partial t}}[/math] must be zero when I really meant [math]\frac{{\partial p}}{{\partial x}}[/math] I have corrected the above.

michel's original post mentioned 'a picture' and asked rather vaguely 'where and when?' What did michel have in mind? A picture would distinguish between northern and southern hemispheres and modern times and Roman times. Many pictures could provide more information, but cameras are not accurate surveying instruments so ascension and declination would have to be estimated in some way from the picture(s). There have been a couple of responses that suggest it is impossible to determine position and time on the planet from astro obs alone. Difficult yes, impossible, no. Accurate knowledge of times is very very helpful, but not essential. Years ago, surveyors used astro time obs to correct their chronometers.

The simple rate equation relies on the reactants being intimately mixed and free to move. This is not the case with solids. Reactions involving solids are multistep. An important step brings the reactants into contact and is often the rate determining step. Here is a guide to solid kinetics. http://www.fhi-berlin.mpg.de/acnew/groups/nanostructures/pages/teaching/pages/teaching__malte_behrens__solid_state_kinetics.pdf

Why not? A given star only transits its zenith over one point at any one time.

What do you mean convert forces? So long as there is no actual movement my lever example provides a 100% 'efficient' force 'convertor'.

No, there is no, best theory. Ideas like most 'delicious ice cream', 'fastest car', 'favourite colour' belong in the junior school playground, not in the study of Science. In Science we look for theories that correspond to observations and are (or should be) prepared to modify these theories in the light of new observations. Atoms bond together because the resultant structure offers the components a lower energy state, or states, than separate existence. Note carefully this may not be the lowest energy state (compare with what I said about best). Just as an architect has many ways to assemble his building, Nature has many ways to assemble atoms. Bonding occurs by interaction between some the electrons in the components. Usually, but not always, it takes place between some of outermost electrons. The octet rule was postulated before the periodic table was fully worked out. At that time they did not know about transition elements. When studying bonding, it is usual to begin with the assembly of a few atoms to form simple molecules. Ionic and covalent bonding are introduced. At some point the idea of delocalisation is introduced. This means that new structures are introduced where the bonding electrons are no longer attributable to individual atoms, they 'belong' to the bonded complex or structure. This introduces a very important form of bonding, metallic bonding, where enormous numbers of atoms and electrons are involved in combining to form a single structure. It also endows the metallic structure with additional properties, not enjoyed by the majority of ionic or covalent compounds. Later minor forms of bonding are studied. These include bonds such as Van der Waals bonds that transient and simply modify existing properties.

Yes +,-,* & / are operators but operators have a pretty wide ranging definition and include operators on numbers. They are like but their definition is even wider than functions since some allow multiple outputs or results, unlike functions. An example here might be the square root extraction operator that can return either the positive or negative square root. Most operators require an argument or operand to operate on but there is even a class of operator that doe not take an argument. These are called nullary or niladic operators. Examples of these are the end, exit and void statements in programming. Other interesting operators are the rotation operator, the negation operator, the complementation operator. The rotation operator has a place in QM.

As I understand it, the longitude problem applies to moving objects such as ships. If you can stand still in one place and watch the stars revolve past then a star almanac will tell you when a certain star transits it's zenith (so long as you know the date which is why I said time range.). So if you can take a series of photos of particular stars at transit you can deduce the (celestial) time. The equation of time relates the local time to universal or celestial time, and thus to longitude. It is at least 35 years since I last did star shots so anyone with better knowledge is welcome to correct me.

Why? Are you thinking of 'the equation of time'?

Do you not think that with photogrammetric techniques the (simultaneous)equations of angles to known fixed points (stars) could be solved without a horizon, if there are sufficient observations?

In principle, yes. The planets are quite variable and only give rough positions. Wiki has a list of the best navigation stars to photograph. http://en.wikipedia.org/wiki/List_of_selected_stars_for_navigation As much information as possible about the time or time range of the photo would be very helpful.

Hi dawg Get hold of How to teach Quantum Physics to your Dog By Chad Orzel That will help you far more than any quickfire answer here.

I would warn against being too loose with terminology. Some mechanical machines are theoretically 100% efficient. There is no theoretical barrier to this. For example a simple lever, in the absence of friction, has an efficiency defined by the ration of the mechanical advantage to the velocity ratio which may be unity. An simple example would be a simple lever. In the real world no mechanical machine is perfect and therefore no real world machine is 100% efficient, but this is due to imperfections, not theoretical barriers. Thermodynamic machines, called heat engines, on the other hand suffer the Carnot limit to their efficiency, which depends upon temperature difference. The difference is that thermodynamic machines convert heat energy into mechanical work and vice versa, whilst mechanical machines put mechanical work in a more convenient form. Please also note that the correct term in current usage for heat pump performance is 'coefficient of performance' or COP, not efficiency.

Looks much better A small point For differentiation with respect to x (d.w.r.t.x) we write [math]\frac{d}{{dx}}(something)[/math] not [math]\frac{{dy}}{{dx}}(something)[/math] So your first line should be [math]\frac{d}{{dx}}({e^{xy}}) - \frac{d}{{dx}}({x^3}) + \frac{d}{{dx}}(3{y^2}) = \frac{d}{{dx}}(1)[/math]

Do you wish to discuss cohesion or do you wish to discuss friction?

Yes that's the right idea for 2(theta)=90 When theta = 0 then the 'triangle' is a vertical diametral line (ST in my diagram)and the chord length is zero. The longest possible chord (PQ in my diagram)is a diameter. A basic property of a circle is that the apex angle of any triangle based on a diameter is 90. So shouldn't your domain be specified by greater than or equal to signs, rather than strictly greater than?

Sorry you are off beam here. for f(x) = A(x)+B(x)+C(x) take logs ln(A(x)+B(x)+C(x)) is not ln A + lnB + lnC. lnA + lbB +lnC is the log of (A times B times C) So have another go.

I don't think you understood my points any more than perhaps I understood yours I have 2 points. 1) Very simply, if you physically have the angle you can draw two lines at this angle. Then by dropping a perpendicular from on to the other and measuring with a ruler (as in Greek geometry) you can calculate the sine by division. 2) But you still don't have an accurate representation of the value of the angle itself, without a protractor. It is interesting as the ancient Greeks were able to divide linear measure very accurately but had no means of dividing circular measure to anything like the same accuracy. This capability came with and spurred on the industrial revolution and enabled many advances. http://www.surveyhistory.org/the_dividing_engine1.htm Before numerical control of machine tools toolroom engineers (and apprentices) used a 'sine bar' to set out angles. http://www.youtube.com/watch?v=bQAhzW4J4qs As regards to computer algorithms the recurrence formula a good starting point is the Cambridge University book Numerical Recipes by Press, Flannery, Teukolsky and Vetterling

Perhaps you will remember the James Bond mnemonic OHMS But I don't see the point of the circle, you can do just as well by drawing a triangle if you have a protractor. If you don't have a protractor, but just have the angle drawn you can still get the sine but you won't know what the angle is then.

Friction and cohesion are entirely different forces with entirely different causes and properties. What was the rest of your post about please?