studiot

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?

Always good to do it yourownself. Hope you also enhanced your knowledge of the properties of circles and how these properties all fit together as a result.

Parabolas are often used in highway engineering particularly for valley curves as they provide a natural slowing down near the bottom. I think, but I am not an expert in fairground rides, that they are used here as well as they give a much better roller coaster rush. I have not seen catenaries used in this respect. Some arched bridges have been constructed with a catenary profile in modern times however.

The rate law is determined experimentally and so is the 'correct' one in the sense that it is what actually happens. However that does not mean that the simple expression obtained from the stochiometric equation is useless. In fact comparison with the experimentally determined rate law tells us when something more complicated, such as intermediate reations, is going on. It also depends whether you insist on whole numbers in your stoichiometry. Take, for instance, a reaction that requires two different molecules A and B to meet and react and form products C and D. Then the rate of reaction will depend upon the presence of both A and B so the assumption is that A + B = C + D (stochiometry) v = k [A] ie is second order. But suppose that the reaction is actually a decomposition 2N2O5 = 4NO2 + O2 experimentally rate equation is v = k[N2O5] not v = k[N2O5]2 This is only first order. But of course the dinitrogen pentoxide molecule does not need to meet another to decompose. The stochiometric equation could in fact be written in terms of a single reactant molecule and fractions for the products.

One thing Imatfaal didn't seem to mention is that his alpha is twice gwiyomi's theta as shown here. Edit just found the hidden meaning in post#5 - spoiler.

You don't seem to have used my suggestion. Your original idea has taken you away from the radius which is a constant for a given circle and given you an expression in terms of the half chord length, which is variable. You really want to do the opposite ie work towards an expression that puts the area in terms of the circle constants and theta alone. Then you can differentiate and set to zero. Here is some help towards this. My final expression can be further simplified, in particular look for a simplification of (1+tan2)

A chord forms the base of your triangle. Angle theta is the angle subtended from the ends of the chord to the circumference to form your triangle at its apex. I suggest you use 2 theta rather than theta and work on half the triangle. Consider the angle subtended to the circumference on the other side of the chord to your isoceles triangle. this also forms an isoceles triangle. The half chord is a side common to both triangles and allows calculation of the division of the diameter.