studiot

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74 downloads of your pdf and no one has challenged it or even commented on it. You have constructed a series of statements without proof. Perhaps you would like to justify and explain what you are talking about since it appears to me to be in direct conflict with conventional wisdom.

http://www.scienceforums.net/topic/85838-why-the-prevalence-of-crackpots-in-physics/

You do not correct the sample mean, only the sample variance. So you always use n-0 to calculate the sample mean. The issue is, as mathematic pointed out, that the mean of a single sampling will probably not match the mean of the whole population exactly. In my example, although the population mean is the most common value amongst the sample means, an individual sample mean equals the population mean in only 1/3 of the possible samples. With only one sample you cannot estimate the variance or standard deviation, unless N = n = 1.

Yes it is mathematically useful in many situations, both classical and quantum. The significance of your first loose definition is that if a function is the same once differentiated (to a constant multiplier) then the solution to a differential equation such as y'' = -ky is an eigenfunction. In other words y is an eigenfunction of this equation. The more general a(x)y''+b(x)y'+c(x)y = -ky Is known as the general eignfunction problem and the area of maths to look up is called Sturm-Liouville theory. S-L theory leads on to adjoint and self adjoint operators as elfmotat has indicated. An important property of eignfunctions is orthogonality which leads them to be linearly independent and forms the basis of useful series solutions. Returning to my first differential equation, this has a familiar general solution y = p*cos(k0.5x) + q*cos(k0.5x) where p and q are constants This may be recognised as the standing wave equation for a stretched string in classical mechanics if the boundary conditions y=0 at x =0 and x=a are added. Using these b. conditions we see that p = 0 and q*sin(k0.5a) = 0 For non trivial solutions since p = 0, q cannot equal 0 and therefore sin(k0.5a) = 0. This happens for the eigenvalue equation sin(k0.5a) = 0, giving the nodes of the standing wave. A similar result can be obtained with complex solutions to the Schrodinger equation in QM

Enthalpy has offered some pretty good comments, to which I would add that Force is measured in Newtons, not kg-metres. Are you thinking of rim or centre drive as it will affect many aspects of the design. In particular Enthalpy has noted the concentration of the mass along the rim, which is fine for rim drive, but the structural requirments for centre drive will require a more massive framework for the 'spokes'. Also the method of drive will be different for centre and rim drive.

Glad you were awake enough to spot the deliberate mistake, now corrected. Hope the rest is helpful, read the attachment in conjunction with the text in the post.

First thing is to get your units straight. You have quoted distances in metric but weight in imperial. Do you know the difference between weight and mass, it is important for rotating bodies. So do you want to work in metric or imperial? Second have you made any effort yourself to solve this question, if so please show where you are stuck.

Take a tip from the Eskimo and build an internal igloo.

Hopefully this is not a real wheel for real passengers, but only an exercise? In which case why is it not in homework help?

OK so what do we actually want to measure when we sample? In other words why do we sample? Well we don't want the actual value for one item. We want a single number that will best represent the whole population. So we want the population average or mean, [math]\mu [/math]. This is given by the formula [math]\mu [/math] = [math]\sum {\frac{{\left( {{X_i}} \right)}}{N}} [/math] That is we add all the individual values, xi up and divide by the number of values in the population. But we also (often) want an idea of the spread of the data. We obtain this as the variance (often reported as the standard deviation, [math]\sigma [/math] or square root of the variance) and given by the formula [math]{\sigma ^2} = \sum {\frac{{\left( {{X_i} - \mu } \right)}}{N}} ^2[/math] That is we add up all the deviations, square, and divide the result by the number of values in the population. But what about the sample? Using upper case letter to denote values from the population, and lower case for values from the sample: If we did the same for only some of the values would be be fairly representing the population mean and variance? Well it turns out that if we took every possible sample of size n < N we find that the average of all the sample means of size n is the same as the population average, [math]\mu [/math], although the sample mean for any particular sample may not be the same as that of the population. But If we take the average variance of all possible samples of size n < N we find is it smaller than the population variance. [math]{\sigma ^2}[/math]. Remembering that we are really interested in the parameter for the population, not the individual sample we find that we can take the sample average as a fair representation of the population average, But, and this is what we want to 'fix' We cannot take the variance of the sample as calculated by the formula [math]{\sigma _s}^2 = \sum {\frac{{\left( {{x_i} - \mu } \right)}}{n}}^2 [/math] as a fair representation of the population variance. Instead of algebra to prove this for all cases the attachment shows a worked example for a very simple case of the population being three numbers {10,20,30} and the sample size being two numbers. So N = 3 and n = 2 It can be seen that the mean of all the sample means is the same as the population mean, but the average variance of all the samples is only half that of the population variance. It can also be seen that bessels correction for this is exactly 2. [math]\frac{n}{{\left( {n - 1} \right)}} = \frac{2}{{\left( {2 - 1} \right)}} = 2[/math] Please also note I have tried to bring out when to use the N or n and when to use (n-1) - we don't use (N-1).

Both the full population and the sample have a mean and a variance (or standard deviation). There is no reason for these two parameters to be the same in both the population and the sample or between two samples unless the sample size equals the whole population. If we take the variance of the sample to be [math]{T^2} = \sum {\frac{{{{\left( {{X_i} - \overline X } \right)}^2}}}{n}} [/math] We would wish it to be equal to the variance of the population, [math]{\sigma ^2}[/math] However some several lines of algebra shows it to be actually equal to [math]{\sigma ^2} - \frac{{{\sigma ^2}}}{n} = \frac{{n - 1}}{n}{\sigma ^2}[/math] So if we 'correct' this deficiency by multiplying this by [math]\frac{n}{{n - 1}}[/math] we obtain the wanted equality. You can see that Bessel's correction is equivalent to using (n-1) instead of n in the calculation of sample variance. Do you really want the algebra proof?

Thank you , SJ, That is what I usually do. But someone told me that this is the 21st century, and I naively thought that computers might have advanced at least to the stage of pencil and paper so I really ought to be able to type it straight in by now. Oh well, back to the chalk and slate.

What is the 5.1mm measurement?

I'm sure I remember seeing a thread somewhen that someone posted how to do it. But I couldn't find it.

Thanks John

Test2 flick toggle switch before posting. This time I copied the table in white under your code box. Last time I copied the code box. n/(n-1) - 2 2.0 5 1.25 10 1.11111 100 1.01010 1000 1.00100 test3 finally got it thanks Acme +1 Number in sample,n Bessel's Correction, n/(n-1) 2 2.0 5 1.25 10 1.11111 100 1.01010 1000 1.00100

test [ta=left][tr][td]n/(n-1)[/td][td]-[/td] [/tr] [tr][td]2[/td][td]2.0[/td][/tr] [tr][td]5[/td][td]1.25[/td][/tr] [tr][td]10[/td][td]1.11111[/td][/tr] [tr][td]100[/td][td]1.01010[/td][/tr] [tr][td]1000[/td][td]1.00100[/td][/tr][/ta]

So what did you do that I didn't?

[ta=left][tr][td] Number in sample, n[/td][td] Bessel’s Correction n/(n-1)[/td] [/tr] [tr][td]2[/td][td]2.0[/td][/tr] [tr][td]5[/td][td]1.25[/td][/tr] [tr][td]10[/td][td]1.11111[/td][/tr] [tr][td]100[/td][td]1.01010[/td][/tr] [tr][td]1000[/td][td]1.00100[/td][/tr][/ta] Well thanks for trying anyway.

So, once again, explain your point. This constant provocative denial of others is beginning to smack of troll.

Here is a word document of what I want to post, if it's any help. bessel's correction.doc

Here is a nice safety oriented video for those playing with Mike's Capacitors asnd Coils. http://www.youtube.com/watch?v=AGXQNLq19FQ

So assume I don't understand, and explain how the link supports your claims. As a matter of fact John Cuthber (and I) referred to carbon dioxide, which is 'heavier' than air. You referred to carbon monoxide which is marginally 'lighter'.

Thank you Acme I see them, can someone tell my how to post a 2 column x 6 row table and how to enter the data?