imatfaal

! Moderator Note Granpa You will not get another warning on any of the below matters 1. Stop being so rude - if you cannot answer questions without appending insults to members then you will endup suspended 2. When a reference is asked for it is not sufficient to say it is obvious, or just post your own work, etc.; members are looking for outside citations 3. Try to accept that you might not have a fast-track to all human wisdom - in the last few threads you have been making out-landish claims with the only back-up being your own work and misunderstood graphs and wikipedia articles Do not respond to this moderation other than via the reporting system

I liked the idea of going for a setup like 3 piece maximum - with base at an offset angle - but just could not visualise it clearly enough in my head and my draughtsmanship skills are not good enough to work on paper for something like that. As I was convinced that such an answer existed I am pleased to see it - and happier that my eventual answer was better I found the same answer for two pieces and worked from there to the five piece answer - rather I had a six piece answer for the same 2 2/3 x 2 2/3 x 2 1/3 and looking at the two piece answer made me realise how I could get that tank with only five pieces (I was stuck on symmetrical sets ---- Two great puzzles TimeSpaceLightForce +1

This is incorrect. T_f is the time for fast ticker at infinite distance at zero gravitational potential - it is not a variable which can be calculated by changing r. T_f is coordinate time - you do not get multiple coordinate times through manipulating this equation. [latex]\frac{t_0}{t_f}=\sqrt{1-\frac{r_s}{r}}[/latex] Around a given black hole (for which mass is the only variable) you get this equation [latex]\frac{t_0}{t_f}=\sqrt{1-\frac{2GM}{c^2r_0}}[/latex] You have two variables and the rest are constants a/o set by the conditions. (I have bolded just for ease of reading) G and c are constants M the mass of the black hole is set by conditions t_f is coordinate time at infinite distance / zero gravitational potential. This is non-negotiable - S'child Metric defines it so; you cannot vary this part. I will now refer to this as the "fastest ticker" - as it is not just fast it is the fastest any clock goes t_0 the time at position of observer - ie proper time r_0 (I have subscripted this to make it obvious) The two variables subscripted 0 are those which vary - you set one and get the other To get the timing difference between two points in a gravity well you have to compare both points separately against the fastest-ticker. 1. I am sorry but there is no doubt that you have been and continue to use one of the most basic equations in this area incorrectly. 2. The equation would not apply anyway as the schwartzchild metric is a massive simplification 3. Even if it did you need to take on board the criticism by Mordred - which is at a much more fundamental level. Please please please - before you come back and claim I am misunderstanding or that you actually mean something slightly different (you will note you have already resiled from the position in your paper) could you just work out the time dilation of the GPS satellites

In my defence in the question there was no mention in op of it being a sensible solution ----- And being very sensible I can get 16.592 cubic feet from five pieces. Making a 2 2/3 x 2 2/3 x 2 1/3 tank Sorry for the iPad graphics. Five pieces - requires six straight cuts to get from 4x8 Reassemble using four welds to form second shape Bend at each of the purple wavy lines and at junction of blue area and clear Weld at two corner joints along one base edge and one centre joint ----- And regarding the units - I agree SI is massively preferable. But in America you buy a 4x8 which makes for easy puzzles - in England you buy a 1220mm x 2440mm which is exactly the same thing but less good for puzzles; and also shows a disturbing lack of confidence and commitment to the metric system

I don't think you can uniquely as there will be two symmetric parabolas with those in common You can find out the vertex from the latus rectum and the focus, as you know that the latus rectum is 4 times the distance of the focus to the vertex You then plug into equation (x-h)^2 = +/-4p (y-k) where p is the distance from focus to vertex and (h,k) are the coordinate of the vertex. The +- gives you two different parabolas one the flipped over version of the other

Here you can find a hand-written scan that I wrote to explain uploading - and the next post in that thread is Michel's explaining how he uses an alternative method http://www.scienceforums.net/topic/73369-uploading-images/#entry732878 nb they seem to have renamed the "Choose..." button to "Browse..."

The first equation in your paper is flat out wrong and you refuse to give a derivation for it. Forget not agreeing with the BB and mirror gradiants etc. Your paper includes this line and Please provide a derivation for this equation - IT IS NOT THE ONE IN WIKIPEDIA FYG - we know Newton's Gravitational Constant to about 5 parts in 100,000 - any equation in which you are multiplying by Newton's Gravitational constant is going to have at least this error. Providing 60 decimal places is just nonsense. In fact anything that relies on a greater accuracy than that of G is something you need to be very careful in handling try a simple jpeg - they normally work. Or upload to image hosting site and use the image tags 11th icon on bottom row of reply options

Best capacity is 130.23 us gallons. You make your tank with these dimensions length = sqrt(32/5)*sqrt(5/3) = 3.266 bredth = sqrt(32/5)*sqrt(5/3) = 3.266 depth= sqrt(32/5)*sqrt(5/12) = 1.633 that gives you a volume of 3.266*3.266*1.633 = 17.42 with a surface area 3.266*3.266 + 3.266*1.633*4 = 32 [mp][/mp] 1. Assume that plan with largest base (ie length*bredth) will also give best volume. Safe to assume - if you think on these lines; whatever depth you choose will maximise volume when area of base is maximised. 2. So what is best rectangular shape to maximise area for any given circumference? a. Assume square base is best but set up variations and see if any are better ie working on the basis of a 1x1 square -> Area = (1-x)(1+x) and vary x to see if Area is ever greater Area = (1-x)(1+x) = 1-x^2 This will always be at maximum when x=0 as -x^2 will always otherwise be negative. Obviously when x=0 the figure is a perfect square. 3. We can set up a similar equation fo volume. Again assume a 1x1x1 five-sided open box - vary the edges and see if we get a greater volume than 1 (always bounded by the assumption in 2 above that best volume will be with square base length = 1+x bredth = 1+x depth = ? We are constrained by surface area - our test five sided 1x1x1 would have a surface area of 5 time 1x1 - ie 5 square units. So to work out depth we take total surface area - subtract that of base (we now have the area of all four vertical sides), divide by four because we need 4 sides( we now have the area of each vertical side), and divide by the length or bredth (and we now have the depth) [latex]depth=\frac{5-(1+x)(1+x)}{4(1+x)}[/latex] Volume obviously equals length x bredth x depth [latex]Volume = (1+x)(1+x)\left(\frac{5-(1+x)(1+x)}{4(1+x)}\right)[/latex] This time you need to differentiate and then set dVolume/dx to zero to get maxima / minima this occurs at [latex]x = -1 +sqrt(5/3) [/latex] Plug this back in and you get length of sqrt(5/3), bredth sqrt(5/3), and depth of sqrt(5/12) 4. That was variation of unit box - to get to a box of surface area of 32 square feet we multiply by sqrt(32/5) which is the size of 5 square sides with total area of 32 square feet That took about 5 times longer to type that it did to work out. Hopefully it is right after all that typing

We really appreciate messages being flagged as spam - but not sure we can do as you ask. Hopefully an Admin can answer. I will say that new threads with dodgy titles - ie the sort that you could flag from the preview screen - are the lowest priority to be flagged; the first thing staff can do is go through the new topics and flag as spammer on the obvious targets. It is the spam that is appended to the end of an existing thread or that has a sensible thread title which is less easy to spot and staff really need to see reported.

You seem to be changing your tune.

you won't get more out than you/nature put in unless you are burning up some other finite resource. If it is not about power (and it shouldn't be) then just use a standard photovoltaic which you can play around with output to get what you need for your electrolysis - no mucking around with the weather and you still get your stuff

[latex] \frac{t_{observer \ at \ r}}{t_{fast \ ticker \ at \ infinite \ distance}}=\sqrt{1-\frac{r_{schwartchild}}{r_{observer}}} [/latex] [latex] \frac{T_{on \ the \ surface \ of \ earth}}{T_{observer \ at \ R}}=\sqrt{1-\frac{R_{schwartzchild}}{R_{observer}}}[/latex] Tada! space backslash space does it for me Mordred On a more serious note - is the OP right to be using Schwartzchild Solution for the EFE in this circumstance. We are not talking about a single heavy mass and the vacuum solution of the space around

Yeah - I don't know how to do spaces in a subscript so I tried to keep it to one word t_fasttickeratinfinitedistance wasn't very intelligible

I have engrossed the subscripts [latex]\frac{t_{observer}}{t_{fastticker}}=\sqrt{1-\frac{r_{schwartchild}}{r_{observer}}}[/latex] This is canonical version from Wikipedia based on the schwartzchild solutions of the EFE [latex]\frac{T_{on the surface}}{T_{Observer_R}}=\sqrt{1-\frac{R_{schwartzchild}}{R_{observer}}}[/latex] This is what you have written Surely you can see they are NOT THE SAME [mp][/mp] Your problem seems to be that you think that the gravitational time dilation equation you have posted gives the time dilation between two arbitrary points at different levels of gravitational potential. It does not. It gives the amount that an observer's clock [proper time] is slower than the clock of someone at zero gravitational potential ie infinite distance [coordinate time]

Hello Martin - just a quick question. In my experience the Aqueous Humour is filling of the portion of the interior of the eyeball between the cornea and the lens (often mentioned as opposed to the vitreous humour which fills the main cavity) ; I don't think this is what you are referring to when you say "aqueous humour" Could you confirm this

http://projects.fivethirtyeight.com/2016-election-forecast/?ex_cid=rrpromo Be afraid - be very afraid

benzoic acid, stearic acid, salt, sand, sawdust, and iron 1st - magnet to remove iron filings 2nd - balloon to get out saw dust. 3rd - nice hot water (over 70 deg C and your stearic acid will liquify and your benzoic acid will dissolve) - stearic acid will form layer on the top, pour off or take off with pipette 4th - filter paper to remove sand 5th - cool water down - Benzoic Acid is has low solubilty in cold water. It will precipitate out and you can refilter for it 6th - boil water off to leave salt just recapping what others said really - but wanted to get straight in my own head. Interested to see what any of our professional chemists say

Please point out in my post where I have mistaken a formula from your work. I have copied a formula and an explanation - and as far as I am concerned it does not match up with any I know It is this bit - please provide a derivation or at least an explanation as to why it looks so like the standard gravitational time dilation formula with a couple of crucial changes. I will remind you that when Strange asked where the above formula came from you tersely directed him at wikipedia. Your formula does not appear in wikipedia. So what is it?

So just use the solar energy to split the water - why get involved in all the loses inherent in such a set up? You will not get out more energy than is put in, by you and by nature, but you can ensure you use the energy nature provides as efficiently as possible. Every machine you put in the loop will lower efficiency

More likely to be Buckie

ok the formula there is gravitational time dilation under schwartzchild metric [latex] \frac{t_0}{t_f}=\sqrt{1-\frac{r_s}{r}} [/latex] t_f being time of fast-ticking at infinite distance t_0 being proper time of slow-ticking observer r_s being the schwartchild radius r being distance (schwartchild coordinate) of the slow-ticking observer what you have is To/Tr = √((1-(Ro/R)) [latex] \frac{T_o}{T_r}=\sqrt{1-\frac{R_o}{R}} [/latex] So the Right Hand sides of both equations are the same - but the Left Hand sides are very different . The correct version in wikipedia is the ratio of time for a slow observer in a gravity well over that of fast-ticker at infinite distance (ie zero gravitational potential); whereas yours is the ratio of time on the earth's surface and time at the arbitrary distance r

I think it is also a matter of time - my worse days of excess were my twenties, my early thirties saw me clean up my act, and my forties have been downright wholesome. I have also transferred my addictions; to get ready for a day's trading/negotiating I would take stimulants, to get me through the day I would chain smoke and caffeinate, to calm down in the evening I would drink, and to have fun I would find other chemical means - and now I cycle. I ride to work and it energizes me, I ride at lunchtime on my turbotrainer to get away from my desk, I ride back to home and it destresses me, and at the weekend I ride my posh bike for pure pleasure. And I spend far too much time on internet fora

Yes - if you take a steady state environment then you will need to provide an input of energy to get the air currents and movement of water which are necessary to the formation; I cannot see how you could possibly collect more energy than you put in. Lots of the energy put in will be lost to you through moving air and water to a state that is at a higher potential. If you are getting less out that you get in ... what is the point?

Not so - in many/most jurisdictions you can no longer patent once the information is in the public domain.