studiot

I know what you are trying to say, my problem is that that which you are trying to say is either self contradictory or contains a hidden 'universal time'. Hiding it in bad mathematics does not make any difference. "One hour inside the box runs out in 30 minutes in the perspective of outside observer, so the time rate is 2 times faster" One hour according to whom ? if you say one hour in the box is equivalent to 1/2 hour outside, or the other way round, that is a transformation not a proportion from one point of view to another. If you say that some observer, who is neither in the box or in the surrounding space finds that he observes one of his hours to last one hour inside the box and 1/2 hour in the surrounding space you have 3 observers not 2. This introduction of an underlying 'absolute time or absolute space is a basic misconception about relativity that founders many an attempt to describe it.

Every scientific discipline has its own reference standards. From your background picture, can I assume you are looking for chemical reference standards ? If so I suggest you ask a moderator to move this from 'other sciences' to chemistry. Please also provide more information as even within the one discipline of chemistry the field of standards is very wide indeed.

I agree that indiscriminate use of the word 'chamber' can lead to false impression. However how big is a chamber ? For instance how big is a vacuole in an amoeba? Is that not. technically a chamber ? And are we not a technical site here ?

Changes with respect to what ? The problem as I see it is that you are offering a rate of change with respect to itself, which is meaningless. using dt'/dt means the change of t' with respect to t. How can this be proportional or anything else ? That is stating a functional relationship which require a common standard to compare by. The simple relationship "distance is proportional to time" has two variables. You only have one.

Well Captain Black chose the lengths that were given as the same to build his replica triangle on. So have you tried doing that with your two equal lengths ?

Well spotted +1

Yes there are plenty of youtube vids of this phenomenon about. Alternatively here is a pdf of an MIT thesis on the mathematics of the phenomenon, plus a full modern laboratory investigation Note the glass deflects, just as I said. https://www.demetraskl.com/pdf/final.pdf I don't seem to have a reply to my comment on Hess's law yet.

Good point Another good point , 'Erasure' is not a feature of thermodymic systems, that I am aware of. Most especially isolated ones since erasure can surely only be effected by an external agent, which by definition in an isolated system does not exist. Your whole post represents some great new thinking, I have just picked out a couple of points to add +1 to. +1 also to Ghideon for his thoughts and colour scheme diagrams.

You have to use a formula that connects angles and lengths. Isoceles/equilateral triangles are the only ones that do this.

Only by direct measurement with a protractor or other measuring equipment.

When I said this I did not construct auxiliary triangles. I did try dropping perpendiculars from T, etc, but very quickly came to the conclusion that this would work but actually involve more work as you do not know the actual length of ST and DV. Therfore this length must cancel out and there are no geometrical theorems where this happens so trigonometric formulae must be involved for calculation. However since the actual length if ST does not matter, the result must be true for any length of ST < SV. So I just chose 3 quite different lengths and drew the figure based on these three quick sketches. Then I measured X in each case and confirmed that it was the same angle.

The glass bowl is not truly rigid. In fact there are no waves in a truly rigid object. The glass bowl is a 3 dimensional version of the two dimensional tuning fork. Yes, I agree that it is even more complicated than I made out since there are actually 2 fluid phases in the glass. That is a very good point. Yes you can excite the air a glass or bottle by the same mechanism that produces the ocean waves - variation in the passage of air across the inteface with the liquid leading to pressure variations that can allow the original lifting of bpdy of water above the equilibrium line.

So SB = SC and KC = KS But how does that help you ? And why have you relabelled your diagram ? Perhaps @Genady will help you. Have you tried doing what I suggested ? It is a nice little problem.

Whilst there are a lot of good points in this and your other recent posts around it, This first sentence is exactly what is isn't. That string is by definition fixed at the ends and so can only excited in between the ends. The glass walls are not fixed but provided the excitment at the extremes. It is also worth realising that there are two orthogonal forces acting - gravity and surface tension - so they can't directly influence each other

I'm sorry to tell you that this is meaningless. You are confusing functional dependence with transformational equivalence.

You are correct that the encoding is 0 for No and 1 for Yes. So the coding therefore depends upon both the order of the questions and the questions themselves, and not the position of the square. So yes the shorter string "11" would identify square 1A. Since @joigus has also observed that the question "Is the coin in square 1A ?" would lead to an even shorter string "1" identifying this square it leads nicely to some further comments. 1) However if the answer was No then there would remain 15 other possible places. So each answer partitions the set into two subsets squares for which the question holds true and squares for which the question is false. 1) I have already observed the set of questions to elicit the identifying code may vary in number. 2) Therefore there exists a minumum set which will identify any given square. 3) The minimum question set that will identify any square is a binay search. This discards half the squares (8) ; Then Half the remaining (4) ; Then again half the remianing (2) ; and finally half the remaining to end with one square. This takes 4 steps and is called a binary search. However this minimum set is not unique and also since the order is important the output string will not be unique either. Such a set might be, and searching for square D4, 1) Is the coin in the upper half ? : No 2) Is the coin in the left hand half of the remainder ? : No 3) Is the coin in the top half of the remainder ? : No 4) Is the coin in the right hand half of the remainder ? : Yes leading to the string "0001" Swapping questions (2) and (4) would lead to the string "0100" Two further comments Firstly about joigus' question what happens if there is only one cell. The actually you would not need to perform any compution since the coin cannot be anywhere else. This brings out the observation I often make about probabilities viz the probabilities 0 and 1 have different properties from any other value inbetween. Secondly since all the information is contained in the questions and answers, the configuration of 'the board' is irrelevant. The cells may be arranged in a line or a ring or scattered.

Does anyone remember the long thread about the mechanism of the 'tin can telephone' (made of two tin cans and a taught lenght of string) we had not too long ago ? Remember also that the mechanism is quite complicated. Here also we have a complicated mechanism. Firstly note the the OP did not say he was vibrating the glass or the glass wall. He said Now that finger introduces a point compressive load on the glass wall, which moves around the glass. The the glass wall acts as a very slender strut under the moving compressive load. On account of the curvature the wall bends slight inward where the finger acts and relaxes elastically again as it moves on. The inward movement slightly decreases the total volume of the container so the water is squeezed slightly upwards locally. Once the finger has moved on and the glass wall has relaxed back to its normal position, the water surface is (again locally) too high up the side of the glass to be in equilibrium under gravity. So gravity pulls it back down again. The falling water moves horizontally away from this point. This produces a rising and falling of the water surface locally, which in turn results in periodic water movements horizontally against surface tension, (which acts horizontally). So these pulse moves across the surface reflecting off the opposite walls creating and maintaining the standing waves observed if the frequency of rotation around the glass is in phase with the resonant frequency of the water. I remember our old A level Physics teacher demonstrating resonance using a 2kg hanging weight struck periodically with a knotted handkerchief. So as you see the phenomenon is quite complicated, separate consideration being needed for all the different forces involved, both horizontal and vertical.

There's no magic in it. Have you abandoned an analytical solution ? As swansont has told you I cannot do your work for you, but I see that you have been trying, even though you have not posted your complete efforts. It is always a good idea to post these as we can then see where you are either stuck or have gone wrong. So here are some analytical hints. 1) Look at triangles SDT and DVT and write out the sine rule for each. 2) Note that DT is common to both these triangles so by comparing (equating) expressions for this common side you can obtain one equation. I obtained one involving the angle have labelled A, the angle you have labelled X and the angle 18. 3) Now look at A, X and 18 from a geometric point of view. They have a geometric relationship in triangle DTV 4) This gives you two equations in 2 unknowns to solve. Let us know how you get on.

Thank you +1. I had been really hoping that someone would look more carefully at my list of 4 questions to reach the answer,because it is not 'optimal' in that my questions will not work in every case. There is, however, an 'optimal' set of 4 questions that will work in every position. This set corresponds to a true 'binary search' where you halve the number of possibilities at each step. This is no accident. A 4 x 4 cell board has 16 = 24 cells. So it takes 4 binary digits to uniquely label each cell. And 16 is the maximum number of cells you can uniquely label (identify) in this way. Does this help ?

Yes, +1 By direct drawing, (ie construction) I very quickly made the result about 10o last night. Sadly this is not enough. As can be seen from my sketch the four angle equations alone are not independent so there are infinitely many solutions, corresponding to the infinite many positions we may place D between S and V. The problem is, however, solvable using the extra information that ST = DV. The fact that we do not need to know its actual length shows that we are not in the ambiguous sine rule case and that the problem is solvable without knowing this length. Genady's constructive solution using isoceles triangles is creative.

Sigh! This paragraph makes no sense and is totally irrelevant to the thread question. How can information entropy decrease ? Happily this question was ably demonstrated by @joigus direct calculation on my example, something you had ample opportunity to perform yourself, but chose not to do. This is now the third thread where you have avoided engaging in direct questions or comments about your posts, the other two being your wave thread and your experiments and information thread. This is a pity because in the case of the latter I supplied a direct answer to your chemistry question ( which you have not acknowledged). My answer here in this thread, without engaging set theory, is that as information entropy and information measure are not exactly inverse functions but reverse ones or contravarient onces, then as information measure increases (which it must do as one progresses through a calculation so you end up with more information (the result) than you started with) information entropy must decrease. I have also pointed out (and once again you have not responded) that one limitation to this would be further times when the calculation is run, you already have the answer.

I'm sorry, this opening post is so full of incorrect data as to belong in the trash can and put behind us all as quickly as possible. Can I suggest that you revisit your zero thread, where you might have asked a serious question but don't seem to have been back to consider the answers?

https://www.bbc.co.uk/news/education-60491719 Whilst I can see the need for at least some knowledge of these subjects, I consider this approach far too heavy handed. How would a modern day Ramanujan ever get into a UK university ? Or how about someone who wishes to follow a Masters in Hairdressing ?

So why did you 'aim' such personally abusive remarks instead of answering the question you still have not answered? So you are now shifting your ground, when I point out that the energy or entropy involved in computation (and it does not matter how long or how much) can vary simply because obtaining the result can take different routes. This is not the case with thermodynamics which is founded on the idea of route independence between defined states.

Thank you for a thoughtful, pleasant and cooperative reply. An example to everyone. +1 I don't think I said that there is no cost to storing (and retrieval) - certainly I did not mean to imply that. Here is an admitedly extreme example. Let us suppose the computation takes 3 hours of running the met office supercomputer and the output result is the temperature at 9 pm in Trafalgar Square at 7.3oC. How much resource does it take to store the result compared to how much it would take to exactly repeat the calculation ? I will certainly look at your references with interest.