studiot

OK this is bit clearer. I suggest that you use a cylindrical chamber rather than flat sided as the corners introduces unwanted effects. The original hovercraft principle was proved using a domestic vacuum cleaner as a blower. You could emulate this experiment. Here are some pics and other useful links to the principle. https://www.google.co.uk/#q=hovercraft+principle This link offers a calculator and analysis of the thrust equations. http://www.hovercraft.org.uk/showthread.php?18868-Fan-Thrust-Calculator 100g sounds rather light, don't forget you have to allow for whatever motor drives your piston although an axial fan may be better.

simong93, You seem to be using different terminology from the rest of us. I'm sure if you explained fully in simple ordinary English what you are trying to achieve someone will be able to help you.

Just to reinforce all that you have been told Vcollision = 30,000 km/sc = 3x107m/s c = Velocity of light is about 3 x108 m/s So v/c is about 0.1, which is the figure swansont mentioned So go check your envelope.

Rather than look for something entitled basic chemistry or basic physics have a look at in the basic materials science section of your local library/college/bookshop. Materials science takes in a wide swath of underlying physics and chemistry without the specialisation or advanced maths you soon get into in the pure sciences. For a US book the book By Callister is good but big. When looking go for something called "materials science" not "mechanics of materials", which is about structural engineering. It is difficult to include bioscience in this but there is a very good book by steven vogel called cats paws and catapaults. This goes through comparing and contrasting natures way of achieving objectives with man made mechanisms for doing the same thing. It is a brilliant starter book and cheap.

A 4D box? yes I think you definitely need to provide more detail to help us understand what you are trying to say.

Yes, the KE should be part of the equations, but have you any figures comparing the relative magnitudes of the KE and anihilation energies? You might be better off considering momentum conservation.

How many dimensions doe this box have? No wonder it levitates. Please try to expand on your question.

Welcome to the technical community, Miranda. You are worried about memorising things and unfortunately your chosen area biology/chemistry has much to memorise. However those sciences are active doing sorts of science and most people find that repeated actions help them remember. They more often they do something, the better they remember. So lots of practice is in order. Do you cook? If so you have another advantage since cooking involves some suprisingly sophisticated chemistry and biology, although you make call the knowledge and skill by different names.

I have tried to tailor this development to include answers to your questions. First a bit of background that may help later. There are many sorts of moments and it is easy to get them mixed up, unfortunately some textbooks do this. The First Moment from a point, P to an axis X-X is the product of some desired quantity, Q, located at P and the shortest or perpendicular distance to the axis. First Moment = Q . d This sounds obvious or pompous, depending upon your point of view, but there is more. In Physics the quantity may be a force, an area, a mass, a pressure or other things. In some disciplines e.g. Statistics or Economics non-physical quantities are employed. We usually don’t bother to include the ‘First’, just calling this moment the moment of the quantity concerned or even just the moment in the case of force. However we find it is useful to define a second type of moment that is called – yes – The Second Moment. This is defined to be the product of Q and the square of the distance d Second Moment = Q . d2 The moment of inertia is one of these second moments and is found when we place mass Mq at P. That is P is a mass point. MOI = Mq . d2, MOI has the units ML2 or Kg (metres)2 Another second moment is called the second moment of area. This is found when an area A is located at P. Thus Second moment of area = A d2 Since this is an area times a distance times a distance it has units L4 or (metres)4 The problem is that many texts, particularly in the field of strength of materials, wrongly call this the moment of inertia so beware. Here is an example of what you might see. The column labelled Moment of Inertia in the table really refers to the second moment of area of a plane section. Note the formula for the “moment of inertia” in the column and the equation at the bottom where you might use this – in bending (flexure) calculations in statics. A body or plane figure (called a lamina) is made up of lots of points P. The relevant first or second moment of the whole body is the sum of all the individual moments of the points P. I mentioned that you need to differentiate and integrate with respect to the appropriate variable. You also need to choose you elemental region appropriately to sum over the whole body. What works for the calculation of one particular quantity may not work for a different quantity. To show this difference I will work through the calculation of volume and MOI of a cylinder of radius a. This will also show why the length is apparently missing from the MOI. To calculate the volume we can consider an elemental disk of thickness dx and sum all the disks from x=0 to x=L (where L is the length) by integrating between these limits. This approach works just fine for the volume and therefore for the mass which we can obtain by multiplying the volume by the density. But when we come to consider moment of inertia there is a problem. In the sketch consider two small elements of face area of the disk, A1 and A2, as shown, such that A1 is distance r1 from X-X and A2 is distance r2. For the volume calculation each elemental area generates a strip (along the length of the cylinder) of the same volume and therefore the same mass. That is the volume does not depend upon r. Vol of strip generated by (A1) = Vol of strip generated by (A2) Mass of strip generated by (A1) = Mass of strip generated by (A2) So all we have to do is sum all the elements A to get the overall face area, which we know is [math]{A_{face}} = \pi {a^2}[/math] However the moment of inertia depends upon r (as r^2) so the contributions from A1 and its strip will be different from the contribution from A2 and its strip. Just integrating along X-X , as we did to find the volume, will not take this into account. So we must choose a more appropriate elemental quantity and variable to integrate. Noting that all the material at distance r from X-X has the same moment of inertia and that this material forms a hollow shell cylinder Note the units of the MOI. This is the version you should use for calculations involving the rotational dynamics of a solid cylinder about its own axis

So I will repeat my question yet again. Object A can move quite independently of object B. So how does the relative velocity help? or if you like what is the gain in introducing object A to monitor the motion of object B against?

OK I owe you a post that has been a long time in the making. It is taking quite a bit to produce this last one but now I know you are still interested in moment of inertia I will complete it.

Check this You should always be able to check your own answers by back substituting your answers into the original equations.

What would your upper limits L and H be? You know they are functions, not constants ?

Thank you for adding information to your original statement. But that does not really answer my question. You must have some origin O to reference you forces Fa and Fb to as well as your postion vectors. Although you have not explicitly stated this I assume they all have a common origin. So you then appear to create an expresion for the relative motion between some particle A and another B, the only connection being the vector origin you have chosen. If we wish to determine the motion of particle B from you equation, What, why and how do we choose particle A, given that A had no influence on the motion of B?

Atomic masses Li :6.94 Fe : 55.85 Al : 26.98 B : 10.81 Au : 196.97 What does 0.0367 times each one of these come to? Notice anything?

Well I left the process of combining equations after I had eliminated one variabele (x). That generated two equations in two unknowns, y and z. a simple method to reduce that to one equation in one variable would be to eliminate (z) by multiplying the second of these by 5 and subtracting from the first. But since this is homework help I left that for you. But do come back for more help as you progress Picking out the right equations to combine is really just a matter of practise.

A little judicious examination of the coefficients in the equations can reduce the work and increase the accuracy of this method.

I don't think you quite mean this.

Have you read any part of the book I recommended?

Perhaps I'm being thick tonight but I fail to see why B has any effect on the motion of A, within the conditions you have stated.

Dogs know when it is dinner time. I seem to remember Chad Orzel had a section on this in his excellent book Teach Quantum Mechanics to your Dog.

There is a certain sense in analysing (1) then (2) then (3) then (4). Note that (4) refers to energy transfer whereas (1) refers to heat only. Failure to make this distinction is a common cause of failure to correctly analyse a thermodynamics problem. Entropy is the ratio of heat transferred to temperature, not the energy. Entropy of what? But no, I would not say that because classical entropy change is defined in terms of cyclic processes and equilibrium states. Can we set entropy aside until the end since it was not mentioned in the question?. The heat engine needs further development to cope with (2) and (3). If you bring bodies A and B (at different temperatures with Ta > Tb) into thermal contact heat will flow directly from A to B. You cannot make this a heat engine ie you cannot extract work from this you have to have an intermediate stage which is why you have to have a working fluid. Incidentally for your issue with (3) Power is the rate of doing work = work/time Note (3) says power, not work. For something that happens infinitely slowly, time = infinity so you are dividing by infinity, which is equivalent to multiplying by zero (in engineering maths not pure maths). So for any finite value of work the power is exactly zero.

You may like to look into 'the LAX heirachy' A good start would be pages 95 to 102 of Cambridge Texts in Applied MAthematics Solitons an Introduction By Drazin and Johnson

What strikes me is that I have made several statements that you haven't asked me about once. For instance I said there are four laws of classical thermodynamics. Each can be stated in several equivalent ways. The most fundamental was introduced after the first law and called the zeroth law. It is the most fundamental because all the other three refer to equilibrium states and there is no point doing this unless you can identify and define an equilibrium state. That is what the zeroth law is about. One version states that heat cannot flow from a colder to a hotter body. So (1) is true but because of the zeroth law, not the second. Any heat pump works like this simple model (there are more complicated ones) There is a working fluid which is cooled by expansion to below the environment temperature. Heat flows from the environment into the working fluid by the zeroth law, cooling the environment and warming the fluid. Because of the zeroth law the fluid cannot become warmer than the environment. The working fluid is then compressed to raise its temperature and pumped to a heat exchanger where it is hotter than the output medium so heat flows from the working fluid to the output medium, again by the zeroth law. The working fluid is pumped back to the environment heat exchanger and expanded to cool it below the environment temperature. This is a cyclic process. At no point is heat transferred directly from a colder to a hotter body.

Was that a freudian slip? As a matter of interest do the students have to pay for this non-teaching?