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Everything posted by studiot
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mad indeed. Take a rod, tube, pipe or other object of circular cross section. Tell me how to measure its radius? The practical world out there works in diameters because that is what a pair of calipers measures.
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Not quite sure what you mean by this. There are many factors in play. Here are some more proceedures from my experience. Given some pieces of steel plate how would you generate a genuine flat surface to any desired degree of accuracy? Given a brick, fresh from a kiln, it takes up moisture and expands slightly over time. This can lead to enormous forces if that brick is restrained as part of a wall. How would you monitor this expansion over time, given that the necessary measurements are of the order of 1/10,000 of an inch? Applied scientists (and researchers) face this sort of issue every day.
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I didn't ask for stress calculations of any description, as they have nothing to do with how much water you need to add to the mix. Yes but that will not give the correct answer, so perforce it is not a correct analysis or model. That is why no-one in the whole world attempts to just add a specific amount of water when mixing structural concrete (or even fence post concrete). What is actually done is a smaller amount of water than calculated by your method is added and then the final amount is 'trimmed' to suit after making workability measurements and then adding some more water. This extra water will be different if every case. Hence the instruction "add water until it is correct" If you would like to know more I will happily tell you, but don't get on your high horse about a subject you are not expert in.
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Thank you for the link. Yes your source offers a specific amount of water. I note he even uses zone2 sand. Have you ever tried this proposed method? Let us suppose you are on site and your concrete is arriving in 6 cu metre mixer lorries. You are in charge so you give the instruction to add exactly 1203 kg of water to each lorry. Now you are pumping this concrete into a large pour and the pump company says they will not pump concrete that has a workability of less than 65mm slump, whilst the resident engineer will reject all concrete that has a workability greater than 75mm slump. Concrete is £80 per cubic metre so each lorryload discarded costs your company £480. Your boss says that any discard will come out of your wages. How much will you loose? PS I have been in this situation on a large bridge deck pour and lost nothing after pouring over 1000 cu metres. But I did not add water your way, or by any mathematical formula. Oh and by the way, all that stuff on plasticity has almost nothing to do with reinforced concrete engineering, even using the plastic design methods available in the codes.
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Humblemum Simulation is generally about process. But underlying both simulation and other forms of modelling is the assumption that the model follows a sufficiently similar set of rules to that being modelled for the product of the processes to be put into correspondence in both systems. In post#141 I posted a process that I maintain cannot be modelled by mathematical formulae alone. I challenged the mathematicians here to offer such formulae if they disagreed. I now challenge you to post a method of simulation for the process of mixing concrete to the desired consistency.
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If you wish to create a new system of thermodynamics based on your own private definitions, you will find it very hard in this world. Better to learn and use the established ones.
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Well I assume you have the equation for the efficiency, which does indeed show that [math]e = 1 - \frac{1}{{{r^{y - 1}}}}[/math] Since the compression ration, r is greater than 1. In the derivation of this equation use is made of the fact that the compressions are isentropic (adiabatic) so that the expressions [math]\frac{{{T_2}}}{{{T_1}}} = \frac{{{T_3}}}{{{T_4}}} = {r^{y - 1}}[/math] hold good. These are used along with the basic definition for efficiency (output/input) to derive the above expression. So the short answer to your question is becasue the two compressions are isentropic (adiabatic).
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Before you can identify your input energy, you must specify your system, system boundary and system process. Input energy is defined as energy that crosses the system boundary during the system process. So if your system is the barrel of fuel, than no, the chemical energy stored in the fuel is not input energy it is internal energy.
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Did you read the referenced article? If so how do you reconcile the solution to the differential equation (where the exponential comes in) with your theory?
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This is a claim but I see no maths to substantiate it. This seems a fairly easy to follow derivation of radioactivity probability. http://www.csupomona.edu/~pbsiegel/bio431/texnotes/chapter2.pdf
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Do magnifying glasses defy entropy by focusing sunlight?
studiot replied to Windevoid's topic in Speculations
Entropy is the ratio of the quantity of heat transferred across a system boundary to the temperature of transfer. -
Do magnifying glasses defy entropy by focusing sunlight?
studiot replied to Windevoid's topic in Speculations
But what does that have to do with entropy? And focused light still travels at the speed of light. -
If it's not a stupid question, how would you move the magnet once you had attached it to the vehicle? Could you not use whatever to pull the car directly?
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Do magnifying glasses defy entropy by focusing sunlight?
studiot replied to Windevoid's topic in Speculations
Defy entropy ? What on earth do you mean? Perhaps you could put some numbers or at least equations to it? -
Sometimes we don't want to reduce the eddy currents. Look up the 'eddy brake' or 'eddy brake disk'. Knowing what enhances the effect can point towards reducing it.
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If you are talking about algebraic identities Consider the two equations 3x2+8x+6 = 2x2+5x+4 3x2+8x+6 = (x2+4x+1) + (2x2+4x+5) The first is only true for certain values of x - This is an equation. The second is true whatever the value of x - This is an identity. So you need to show that the condition is true for all x.
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I do not have access to this book now, but try here Charles Taylor : Ancient and Modern Geometry of Conics (Cambridge 1881) contains much history of the subject. The original book was by John Wallis in about 1661
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As I said I was not available over the bank holiday but can carry on now. The problem with air bubbles is that they presumabably lead to a homogenous material. Entrained bubbles will also lower the susceptibility in the flux direction. The point I have been trying to emphasise is that the eddy currents act in a particular direction, so you only want to change the properties in that direction. I also noted that eddy currents are circulatory in nature. The reverse emf they generate depends upon the area of the circle. Again entrained bubbles would not reduce that area in the way that complete separation by lamination provides. Finally in any material where the emf is induced by motion in a steady field, rather than a changing field, the emf is induced in all the moving material immersed in the field. You need some part of the material that is not immersed in the field for the return half of the circulatory current to flow.
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I do believe that this relationship was known to the ancientGreeks. Pappus wrote The four books of Euclid's 'Conics' were completed by Appolonius, who added four more and gave us eight books of Conics. However Euclid's books were lost. Aristaeus wrote five further books of the Conics. These books contained all the knowledge to prove geometrically the relationship ie equivalent statements, so perhaps it was never explicity stated, it was used however.
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Hello gwyiomi, since this is homework, I was looking for your start on the problem, or at least your thoughts on how to start, but couldn't find them. Hint: Can you rewrite the equation of the line in a more useful manner?
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Why do so many posters want to put the boot in ?
studiot replied to studiot's topic in Suggestions, Comments and Support
I think we are loosing the point that nobody's theory is perfect (even Einstein's). Every theory is only a model and the wise man knows when to use it and when to look for something better. He does not automatically say "The theory did not work out 'here' so should be discarded for all other purposes, as so many posters do."