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August 23

quarter car

Can anybody give me reference (preferably on the web)to the method of solution of differential equation of a car suspension(a quarter car) for finding International Roughness index of a road profile? — Preceding unsigned comment added by Amrahs (talk • contribs) 18:24, 23 August 2012 (UTC)[reply]

Wouldn't neglecting the effect of the other 3 wheels on the car as a whole lead to some rather inaccurate results ? After all, the jounce the occupants feel is not simply the jounce at 1/4th of the car multiplied by 4, they may either cancel or reinforce each other, as with waves. StuRat (talk) 05:09, 24 August 2012 (UTC)[reply]

Personal attack

The following discussion has been closed. Please do not modify it.

StuRat: Your initial thought should have been "I don't know the answer to this." And so your reply should have been non-existent. Please stop replying for the sake of it. You're the same on all of the other reference desks. — Fly by Night (talk) 00:10, 25 August 2012 (UTC)[reply]

If you disagree with my statements, you should indicate why, with sources, here. However, your last 2 sentences are a violation of WP:NPA. In any case, this is not the appropriate place for such statements. Shall we move them to our talk pages ? StuRat (talk) 04:50, 25 August 2012 (UTC) [reply]

Sampling for proportion: proportion of what?

I need to take a quality-control sample for a case-reading study. A random sample of about 600 cases will be drawn and read, and I need a sample of those to be read by a second reader. The case-reading instrument has approximately 100 questions—the final number hasn't been determined yet. The response choices mostly form ordinal scales, although a few will probably be nominal, hence I'm sampling for proportion.

Everything I can find on sampling for proportion only looks at sampling the proportions for a single characteristic of interest. I have 100+ of them. How do I approach this problem?

Many thanks,

Henry — Preceding unsigned comment added by HIinNYC (talk • contribs) 21:29, 23 August 2012 (UTC)[reply]

Why are you doing this? What do you need to know? Looie496 (talk) 04:21, 24 August 2012 (UTC)[reply]

Every nonfinitely generated abelian group must contain one of...

Is there any list of relatively simple nonfinitely generated abelian groups, one of which must be contained in any arbitrary nonfinitely generated abelian group? For instance, such a list might include all subgroups of $\mathbb {Q}$ , and all direct sums of countably many nontrivial quotients of $\mathbb {Z}$ . --72.179.51.218 (talk) 23:41, 23 August 2012 (UTC)[reply]

An attempt to answer my own question. I believe the following is such a list. Let

A

be an abelian group:

Case 1:

A

contains an infinite sequence of divisible elements, ie, a sequence

x_{1},x_{2},\cdots

such that

\langle x_{i}\rangle \subsetneq \langle x_{i+1}\rangle

. Then

\langle x_{1},x_{2},\cdots \rangle

is isomorphic to a subgroup of

\mathbb {Q}

or

\mathbb {Q} /\langle 1\rangle

.

Case 2:

A

does not contain such an infinite sequence, and its torsion group is not finitely generated. [editted] Then by CRT, only finitely many primes occur as the order of some element, because otherwise every finite order element is divisible. One of those primes must occur infinitely often, because the elements of order

p^{i}

generate the torsion subgroup, again by CRT. It takes a bit more reasoning, using the classification of finitely generated abelian groups, to show that it follows that there are infinitely many elements of order

p

. Then countably infinitely many such elements generate a vector space over

\mathbb {F} _{p}

of countably infinite dimension - ie,

\left(\mathbb {Z} /p\mathbb {Z} \right)^{\aleph _{0}}

.

Case 3: Otherwise, must

A

necessarily contain a direct sum of countably many copies of

\mathbb {Z}

?

--72.179.51.218 (talk) 00:37, 24 August 2012 (UTC)[reply]

I don't think either of the two classes you mentioned would fall on that list, because you can give examples of non-finitely generated abelian groups whose elements all have finite order. On the other end of the spectrum, you can make such a group whose elements all have infinite order, too! Rschwieb (talk) 20:52, 24 August 2012 (UTC)[reply]

I listed groups in both categories... --72.179.51.218 (talk) 22:36, 24 August 2012 (UTC)[reply]

Oh, I think you're misunderstanding my premise. I'm not asking for a list of groups, all of which must occur as a subgroup of an arbitrary nonfinitely generated abelian group. Just a list such that every nonfinitely generated abelian group contains one of them. --72.179.51.218 (talk) 22:42, 24 August 2012 (UTC)[reply]