Carrock

I don't think using the Schwartzchild radius of a BH as a reference for gravitational potential is helpful here. You may not intend this, but you seem to be suggesting that masses cause nearby spacetime ( and matter ) to have greater energy than distant spacetime ( and matter ) while the usual idea is that on average the mass-energy of matter etc is exactly or nearly equal to its negative gravitational mass-energy. An analogy for the local high energy of an accretion disk would be an Apollo capsule flying through the atmosphere at 25000 knots after falling from the moon. Locally, a vast amount of kinetic energy seems to have been created from nothing; the simplest way to avoid this is to say that the capsule's increased kinetic energy is balanced by its decreased gravitational energy. If a BH created a nearby positive energy/density region that energy would be absorbed by the BH causing exponential mass-energy creation.

I just went outdoors to check the temperature (14 Centigrade). Is it a warm day?

Tech savvy criminals will presumably now abandon this iPhone and use a secure method of encryption. A law against secure storage of data in America and elsewhere would generate a lot of new jobs. Perhaps it's time to build on the successes of the war against drugs and the war against terror with a new war against security.

This is an instance when our perceptions mislead us because accuracy in normal situations is unhelpful, even dangerous. If you see a running lion or a fast car in the distance, not heading towards you, it seems slow and not much of a threat. If it's getting any bigger, it's worth some extra attention. You give most of your attention to more important things. Anyone who's tried to approach a nervous wild animal will be particularly aware of this attitude. Think of walking along a quiet road near a busy road and giving equal attention to each of the dozens of cars on that road and the one car on the quiet road which happens to be heading straight towards you.

One of Richard Feynman's tall tales was about a discussion with philosophers about whether an electron was a thing. To clarify their definition of a thing, he asked if a brick was a thing. They never did get back to the electron.....

Creating a spacecraft with a 100 microFarad capacitance at 100kV is a few orders of magnitude beyond current technology.....

Solar panels are basically diodes optimised to produce current from incident photons. So a dark panel in series with lit panels is basically a forward biased diode. It will pass current with a small voltage drop ( eg about 0.6v for silicon).

I only intended the reference once, but used cut and paste without checking. As you expect me to read your references without specifying which part, I thought you would be happy to do the same or at least google "dimensional analysis". I looked at a few references and didn't find any suitable examples. I only use it as an informal quick check of equations and I thought you would want a formal description. I was taught it in an off the syllabus end of term maths class at school and don't know or care if I'm using it correctly as long as it gets results. This is problematic when I refer to it here. Informal worked examples of dimensional analysis using all my knowledge:-) .. There may be errors. [latex]dU=-pdf[/latex] Since dy/dt in this context is equivalent to y/t, multiply both sides by t; set sign positive (also ok). then L.H.S. = energy R.H.S.= pressure x force In S.I. units L.H.S. = kg*m2*s-2 R.H.S. = (kg*m-1*s-2) * (kg*m*s−2) = kg2*s-4 The dimensional inequality is sufficient to show the equation is not valid. Also useful: exponents must be dimensionless. Worked example: [latex]V=V_0 e^{-\frac{t}{RC}}[/latex] [latex]\frac{t}{RC}[/latex] ie [latex]\frac {time}{resistance*capacitance}[/latex] must be dimensionless. In S.I. units [latex]\frac {time}{resistance*capacitance} = \frac{s}{(kg*m^{2}*s^{-3}*A^{-2})*(kg^{-1}*m^{-2}*s^4*A^2)} =1[/latex] Exponent is dimensionless. It's often possible to find dimensional errors by inspection if the equation is simplified e.g. [latex]\dot{r}=-3\rho \frac{\dot{r}}{r}[/latex] solving for [latex]\rho [/latex]: [latex]\rho = -\frac{r}{3}[/latex] which 'looks' as well as is wrong. Perhaps you meant [latex]\dot{\rho}=-3\rho \frac{\dot{r}}{r}[/latex]

This post was quoted on another thread by Mordred Beyond a demonstration that the 'right' answer need not depend on correct equations I don't see its relevance there so I'm responding to it here before looking at the rest of his post. It's quite simple to spot many of the errors but I haven't made much attempt to spot the more subtle ones. dv=-pdF fails dimensional analysis. Implicitly requiring the volume to be cubic or irregular is problematic initially and would require your next correct equation to be changed. Perhaps you meant [latex]V={\frac{4}{3}}\pi r^3[/latex]? Why bother defining it when you don't bother defining most of your other terms? [latex]\rho=-3(\rho+p)\frac{\dot{r}}{r}[/latex] fails dimensional analysis. [latex]\rho=\frac{M}{4/3\pi r^3}[/latex] I expect it's OK but I've never been that strong on explicit and implicit operator precedence. Why not put [latex]\rho=\frac{3M}{4\pi r^3} [/latex] or at least avoid using different styles in the same equation [latex]\rho=\frac{M}{\frac{4}{3}\pi r^3} [/latex]? [latex]\rho=\frac{dp}{dr}[/latex] looks wrong but with no context it's not obvious. [latex]\dot{r}=3\rho \frac{\dot{r}}{r}[/latex] fails dimensional analysis. [latex]c=f\lambda=\frac{2L}{N}[/latex] fails dimensional analysis. With so many errors and so few definitions and intermediate steps it's very hard to follow this post. To quote the expert I only used dimensional analysis and only for the obvious errors so there may well be other errors. (Having dimensional homogeneity is a necessary but not sufficient requirement for an equation's validity.) The thought of having to fully check every relevant equation in the rest of Mordred's post on the other thread is rather daunting and I am wondering if there's any point. Halfway through this post I lost the will to post but I eventually decided to finish it so I wouldn't have wasted my time completely. If the errors here are not fixed and much more detail provided I would recommend reading these cosmology lectures which cover extremely similar ground in a much clearer and more accurate way. Somewhere along the way I forgot that my objection was the inappropriate use of classical physics to derive cooling of the CMBR. Using pv=nRt is perfectly valid. So in terms of admitted mistakes Mordred is winning with fewer than me.

I wasn't criticising Mordred's results, simply his use of part of a classical theory to prove results inconsistent with that theory. I suspect we ascribe different meanings to the phrase 'Mordred's analysis works.' Some gas laws are used in cosmology. I didn't see pV=nRt being used in your offsite references as a well known explanation of the cooling of the universe. Quote please. 'Energy is conserved' : can you demonstrate that your definition of energy conservation is the same as that used in pV=nRt? See eg. http://www.preposterousuniverse.com/blog/2010/02/22/energy-is-not-conserved/ Irritating isn't it when someone posts bare links because they're too lazy to quote or can't/won't present a possibly refutable argument? 'pressure in the FLRW metric is the direct result of particle to particle interactions. Reference and quote please, not just another set of offsite links. If you believe the above, presumably you think there was a sudden drop in pressure when the CMBR decoupled from matter, or the CMBR is still in equilibrium ie its photons are still interacting. Which is it? 'This is termed an adiabatic gas in thermodynamic terms.' CMBR cooling probably is adiabatic. From your reference https://en.m.wikipedia.org/wiki/Equation_of_state_(cosmology): As both equations agree 'that particles with higher kinetic energy exert greater pressure' and they are closely related I presume you are saying that [latex]w = \frac{p}{\rho}[/latex] and pV=nRt can be used interchangably in cosmology. I'm familiar with more than one scale factor so I don't know which you're referring to. Your scale factor is not referred to in your reference; perhaps you're confusing the symbol t(time) with T(temperature). Reading and trying to understand your references might help you avoid such errors. I'm guessing your scale factor is the first referred to here. I expect you can easily prove/refute the above using pV=nRt but here I demonstrate my close mindedness by saying your analysis will be incorrect.

This may be a little misleading. Classical physics does not claim that increasing volume violates energy conservation, as you seem to imply. Assume a sealed container of gas. When you slowly expand the container the gas does work on the container walls. Its temperature goes down because some energy has been removed from the gas in the form of work etc. Energy has not been destroyed. Classically if you release some gas into vacuum, the gas is not in equilibrium and the volume containing all the gas increases indefinitely. The total energy of the gas does not decrease as the volume increases. Black body radiation refers to radiation from a body in thermal equilibrium. As the CMBR is not in thermal equilibrium in the sense required for equilibrium thermodynamics to be valid (which it has not been at least since ~400,000 years after the B.B.), the stretching of wavelength is due to the spatial expansion of the universe and nothing to do with near equilibrium thermodynamics. Indeed this creation of space and wavelength stretching is in violation of classical physics. As it happens, the stretched CMBR has the same spectrum as radiation from a 2.7K black body, with a few imperfections. This is pretty consistent with the CMBR being radiation from a black body at 2.7k but there is a massive amount of evidence inconsistent with this view. It might be less confusing to think of the CMBR having a colour temperature of 2.7K much like a cool running LED lamp might be described as having a colour temperature of 3,000K. ie colour temperature is a description of a spectrum, not the temperature of the source.

You clearly regard a gun's potential to kill the people it's aimed at as the primary, perhaps only, measure of its safety. I'd agree to an extent. However, the widespread view that guns improve personal safety, the general absence of a requirement for competence and the idea that safety is measured by the ability to shoot the intended target are mitigating circumstances. There is no requirement, in general, for competence in a gun owner, so in many cases even a 'safe' gun will kill apparently innocent bystanders rather than suspects. The number of shots that can be fired is important, since the more shots, the better the chance of killing the suspect. A perfect crime if you're careful and a bit lucky: I've seen a few 'true crime' stories where a spouse kills several partners 'accidentally' before the police get suspicious enough to look for evidence of murder. Even then, abysmal stupidity by the perp or dubious evidence is often required for a conviction. The most powerful, lethal gun (with a light trigger action and easily disabled safety catch) is therefor the 'safest'.

More dangerous for the user certainly. Even the most law abiding citizen is potentially dangerous and may get you in legal trouble after you fail to kill him/her with an inadequately lethal gun, even if you honestly believed (s)he was intending to attempt to kill you.

Because the earth wobbles ie precesses, Polaris will not be the pole star indefinitely. From https://en.wikipedia.org/wiki/Thuban

Exponential spatial expansion forever is not compatible with known or AFAIK speculative physics. There would be [latex]{2}^{\aleph_{0}}[/latex] finite units of space ie as many finite units as there are zero volume points in the universe, which is generally considered impossible. This issue can be avoided by saying that that at indefinitely distant past and future times the universe is not expanding, but you still have to explain why expansion started and eventually stopped.

But you would be wrong within currently understood physics, so that would be a bad idea. Seems I was wrong about the meaning of .

You could also say that you and the ball are stationary and the ball is pulled outwards by gravitational force caused by the relativistic effects of the universe rotating around you.

This maths is rather beyond me, which is why I stopped posting in these topics. However, from http://web.mat.bham.ac.uk/R.W.Kaye/seqser/density.html and some other pages, I couldn't find any any indication that the set of integers is not dense in the same sense that the rational numbers is dense. e.g.

Small correction maybe. Surely the rationals must be discrete or they'd be uncountable and could not be mapped to the integers.

No, but I said moved, which your latest post indicates is impossible, not removed. Overvoltage can destroy resistors though I'd not expect it if the 110V rail is always between 70 and 150V (which I suspect it isn't). C8 ensures that most of the A.C. voltage transients appear across R25/R26. C8 isn't needed to protect the inverter but it may be needed for reasons you know about but I don't. Based on the limited information you've provided I'd suggest: Remove C8 and replace R25/R26 with either one 110 ohm 1 watt resistor ( for its higher voltage rating) or with two one watt 220 ohm resistors in parallel. There are many possible reasons this won't work but I'm done with guesswork and this topic. I hope this helped.

A few points. If the converter is switch mode I doubt there's significant equivalent input capacitance. Is the dc input fairly constant? eg rectifier+capacitor? (But rectifier+capacitor drifts up to peak input voltage - 150V - if there's no load eg prior to closing DC power switch.) If so C8 isn't needed or can be replaced by a smaller value. It's probably taking nearly all the surge current rather than the 110Vdc to 24Vdc converter. It may then be better to switch the power source for the 110V D.C. input if that is possible rather than the D.C. output. If the input is rectified unfiltered A.C. C8,R25 and R26 should be moved. I think that would be a safe approximation.

This could be good, depending what you want to use the pulse duration length for. I suspect your application has many similar solutions by other people and you could adapt your circuit in that light. eg is it really necessary to deliver 0.76J at 50Watts?

You're relying on the non repetitive surge rating of the resistor, which is a function of peak power and energy of the surge. You may also be exceeding the smaller resistor's voltage rating. These things depend on the resistor design; you'd need the manufactuter's data. The whole resistor doesn't have to warm up for the resistive part to burn out; a fuse burnt out by a surge typically remains cold while it may be warm if it blows through a continuous slight overload. If you're using a 10W wire wound resistor I'd expect the thermal inertia of the wire to cope with a 3 1/2 joule 50W peak pulse. Otherwise, without more information the only reliable solution is a 50W resistor. [edit] maybe 25W if two resistors[/edit]

Yes. When n→∞, it has to start from 1, then +1 then +1 ...... At each step, n is a finite number. In fact, n is not kept finite, but cannot be infinite, that is, never n=aleph_0. So, real number cannot be mirrored by digital numbers. I will try another approach. I assume you accept that there are [latex]{\aleph_{0}}[/latex] elements in the infinite integer series 1,2,3..... ie the series does not end at some arbitrarily large finite number. If not, I'm done. Rather than count n by starting from 1, avoid that objection by simply constructing figure 3 in the same way as you construct the infinite integer series, with each distinct n set equal to a distinct positive integer. You don't have to count or iterate or step through every or any value of n; constructing either the infinite integer series or figure 3 does not require you to 'count' every member. If you consider the number of rows generated by different values of n to be finite, you have to explain how n cannot be set equal to some members of the infinite integer series 1,2,3..... If this proof I've outlined here and earlier in thread, that there are [latex]{2}^{\aleph_{0}}[/latex] ie [latex]{\aleph_{1}}[/latex] distinct binary fractions is valid, your contradicting proofs must be invalid.

Bertrand Russell opposed this view saying, (approximately) that such a series can be defined and used. There's a good discussion of this issue, which I won't/can't attempt to argue, on page 29 of Philosophy of Science Vol. 32, No. 1 (Jan., 1965). (free registration on site required to read it) Even if you don't accept this, I think there is another flaw. If n can't ever reach aleph-null, then you similarly can't ever reach even an (aleph-null)th member of the above counts ie you can only make an unbounded finite count. If you argue that you can calculate what the result would be if you could count up to aleph-null, then are you not accepting you can similarly calculate the result of n reaching aleph-null?