Subject: Re: From strong pseudoprimality to primality proving
From: rgep@dpmms.cam.ac.uk (Richard Pinch)
Date: 15 Nov 1996 14:58:23 GMT
In article ,
"Vyacheslav Lukyanov" writes:
|> It is known due to E. Bach that assuming generalized Riemann hypothesis (GRH)
|> a number n is prime iff it passes the Miller-Rabin test to all bases less
|> than 2(log n)^2 + 1. If one does not assume the GRH then a result of
|> D.A. Burgess implies that testing to bases up to n^{0.134} is sufficient.
|> Do somebody have an idea about the result and how does it imply the subject?
The Euler-Jacobi criterion is that b^((n-1)/2) should equal (b|n) mod n, where
(b|n) is the Jacobi symbol. The set of b mod n which satisfy this criterion is
a subgroup of the multiplicative group modulo n, and a proper subgroup if n is
composite. The Bach result is that no proper subgroup can contain all the
numbers from 1 to 2(log n)^2. The Burgess result is that you can find b which
is a non-square modulo the smallest prime factor p of n, and a square modulo
the others, so that it is bound to fail the criterion: the result followed
from an estimate on the least non-square modulo p.
--
Richard Pinch Queens' College, Cambridge
rgep@cam.ac.uk http://www.dpmms.cam.ac.uk/~rgep
Subject: Re: Problema de comunicacion !
From: Alexander Anderson
Date: Fri, 15 Nov 1996 14:15:24 +0000
In article <568bok$qtl@hp.fciencias.unam.mx>, Mario Ramos Andrade
writes
>es
>
>Armio
>Prof. Mario Ramos Andrade
>____________________________________________
>
>Hay probablemente más puntos en el espacio y más momentos en el tiempo que
>números finitos.
>BERTRAND RUSSELL
>(1956)
>____________________________________________
>
>Para mi el tiempo se da en la memoria, ciertamente, y sólo es posible como
>registro de lo que recordamos.
>JORGE LUIS BORES
>(1989)
>____________________________________________
>
>¿ Como transmitir a los otros el infinito Aleph, que mi temerosa memoria apenas
>abarca ?
>JORGE LUIS BORGES
>(1988)
Mathematica tem problemas com verdades nao provabel.
Mas o Realidade e o Realidade...
Communicacao, finito, es entre quem e quem?
Nao tem communicaco (finito) ...nao tem universo.
Sandy
--
// Alexander Anderson Computer Systems Student //
// sandy@almide.demon.co.uk Middlesex University //
// Home Fone: +44 (0) 171-794-4543 Bounds Green //
// http://www.mdx.ac.uk/~alexander9 London U.K. //
Subject: Re: Fields ?
From: Jan Rosenzweig
Date: Fri, 15 Nov 1996 10:45:56 -0500
Philippe Langevin wrote:
>
> Why do the mathematicians call a FIELD (like R and C) a field in
> english,
> and UN CORPS in french ?
Isn't une corps a division ring, i.e. you do not assume
multiplication to be commutative?
--
Jan Rosenzweig
e-mail: rosen@math.mcgill.ca
office: home:
Department of Mathematics and Statistics 539 Rue Prince Arthur O.
Burnside Hall, room 1132, mbox F-10 Montreal
805 Rue Sherbrooke O. Quebec H2X 1T6
Montreal, Quebec H3A 2K6
"It is unworthy of excellent men to lose hours, like slaves, in
the labors of calculation" ..... Leibnitz
Subject: Re: Mission Impossible: Can probability=0 events occur?
From: jyoung@ultra0.rdrc.rpi.edu (Jeffrey A. Young)
Date: 13 Nov 1996 16:51:55 -0500
In article <569p9l$rlj@gap.cco.caltech.edu>,
Ilias Kastanas wrote:
>In article <567rpd$nh@ultra0.rdrc.rpi.edu>,
>Jeffrey A. Young wrote:
>>In article <561do5$2ui@gap.cco.caltech.edu>,
>>Ilias Kastanas wrote:
>>>In article <5607lo$rvj@ultra0.rdrc.rpi.edu>,
>>>Jeffrey A. Young wrote:
>>>>In article <55tvo6$un@sun001.spd.dsccc.com>,
>>>>Mike McCarty wrote:
>>>>>In article <55lpte$osm@ultra0.rdrc.rpi.edu>,
>>>>>Jeffrey A. Young wrote:
>>>
>>>>Not at all. Every event (sequence) has probability zero.
>>>>Only the individual events you can specify are impossible.
>>>>
>>>>>You are also pretty fuzzy about "fully specify". What does that mean?
>>>>
>>>>Well substitute "identify explicitly as an individual event before the trial
>>>>occurs" then. Obviously there are uncountably many events in the population
>>>>and only a countable number that anyone could specify individually
>>>>beforehand.
>>>>
>>>>>If I specify the class of events which can occur, then I can certainly
>>>>>predict what the outcome of the trial will be. It will be one of the events.
>>>>
>>>>But you cannot specify all the events in the class individually.
>>>>That is the difference.
>>>
>>> This notion of 'impossible' does not seem to make sense. You are
>>> granting the feasibility of infinitely many trials. In any single trial
>>> both H and T are possible outcomes. So how can any infinite sequence of
>>> H's and T's be "impossible"?
>>
>>How can you claim to know anything about what happens at infinity?
>
> You made such a claim in the first place.
No, my claims are all in regard to what *cannot* happen at infinity,
while yours are in regard to what *can* happen, a crucial distinction
in this argument.
>>>What difference does it make whether it can be "fully specified" or not?
>>
>>It was intended as one possible counter-argument.
>
> To what? Sorry if I missed an argument of yours, but I didn't see
> any; just the assertion that some sequences are impossible.
I believe it was Mr. Ullrich who was trying to assert that we had to either
1) accept that all sequences were impossible (because probability was zero), or
2) accept that all sequences are in fact possible, although probability is zero.
I refute the exclusivity of the above by showing a possible (!) middle ground.
>>> Any one of H, HH, HHH ... is surely possible; what does it mean,
>>> then, to say "HHHH... is impossible"? Same for any infinite sequence;
>>> all its finite initial segments are possible. Stating a priori that even
>>> one particular infinite sequence is "impossible" contradicts the assumption
>>> about single trials; it sounds like a conspiracy theory, or animism.
>>
>>Uhhh, I don't see anything about "infinity" in the assumption
>>about single trials. Aren't you using other assumptions as well?
>
> The one I stated above, that infinitely many single trials are possi-
> ble. You apparently agree with that. So why is it impossible for all to
> come out H?
>
> In another posting you stated your claim is _consistent_. That is a
> retreat from saying it is true (parallel postulate, etc)... but I'm afraid
> it isn't even that. It is fine to define a uniform probability on some
> subset of [0, 1], and have everything else a priori impossible. But when
> you consider infinite sequences of H and T it is plain logical error to
> assert that some are not there.
Not where? Nowhere do I assert that some are not in the population,
only that some are impossible to produce. Are you asserting that every
member of a population is necessarily the possible outcome of a trial
against that population? That's hardly true. Perhaps you are trying
to assert that all impossible results are easily identified and eliminated
from a population before the trial? I doubt it, but maybe you could show
me some of the "plain logic" that will convince me. Apologies if you
meant something else by "not there" (but please explain).
Jeff
Subject: Re: Decision tree problem
From: Richard H Gould
Date: Fri, 15 Nov 1996 13:43:10 +0000
In article <328AC977.156F@sagelink.com>, rick
writes
>Suppose you have 12 coins. 1 is lighter or heavier than the rest.
>Determine the bad coin and whether it is heavy or light using at most 3
>weighings. Your descion tree should have 24 outcomes at the bottom.
>
I first saw this in an Oxford scholarship paper 35 years ago and it
remains a favourite; however, it is too good a problem to spoil, so I
won't. As stated the problem is not quite fully specified. The
weighings are carried out on a balance by weighing coins against coins.
The clue I will give to the full problem, as stated, is what I call the
'3 coin lemma', and was also given in the original question:
How many such weighings does it take to distinguish one out of 3 coins,
known to be lighter than the other two?.
--
Richard H Gould
rhgould@gocomp.demon.co.uk
Subject: Re: Need help on an stats/prob question
From: touhey@aol.com
Date: 15 Nov 1996 16:42:55 GMT
The question was:
> Suppose the average family income of an area is $10,000.
> a) Find and upper bound for the percentage of families with incomes
> over $50,000.
> b) Find a better upper bound if it is known that the standard
> deviation of incomes is $8,000.
> I assume that some kind of distribution must be assumed.
> Thanks for your time.
HINTS:
For part a) the smallest possible family income is $0.
The mean income is $10,000. What percentage of families could possibly
make over $50,000 assuming everyone else is making $0.
For part b) use Chebyshev's ( Tschebeysheff's ) Theorem. And no you don't
need to know the distribution.
Subject: Re: ZEROS of the 13th degree polynomials
From: lange@gpu4.srv.ualberta.ca (U Lange)
Date: 15 Nov 1996 17:04:06 GMT
tleko@aol.com wrote:
: In article <56fjqj$c6q@pulp.ucs.ualberta.ca> lange@gpu5.srv.ualberta.ca
: wrote:
: >
: > Are you satisfied by the example below? Your MATLAB-programs are
: > pointless.
: > roots([1 2 3 4 5 6 7 8 9 10 11 12 13])
:
: Not yet. It is too early to argue about the roundoff errors in the
: two programs while we have only 13 degree polynomials to compare
: minor discrepancies.
: You can write p=[1 2 3 .... 200] not to have something to compare
: with.
Welcome to the trap ;-)
Why should it be "too early" to talk about round-off errors? "roots()"
determines the zeros by calculating the eigenvalues of the companion
matrix of the polynomial. Hence the round-off error influence simply
depends on the condition number of this matrix, which can be easily
determined (e.g. using MATLABs "cond()" or "condest()") for a particular
polynomial of degree 200. That`s all basic numerical analysis.
_You_ have no clue how the round-off error spoils the results of _your_
messy code (BTW: What`s the round-off error influence in a command
like "contour()" ;-). But people who are doing these things _seriously_,
like the ones who implemented MATLAB, know what they are doing.
--
Ulrich Lange Dept. of Chemical Engineering
University of Alberta
lange@gpu.srv.ualberta.ca Edmonton, Alberta, T6G 2G6, Canada
Subject: Re: Why is 0! = 1 ?
From: Jon Haugsand
Date: 15 Nov 1996 18:29:22 +0100
Cyc0@gothenburg.irc.pp.se (Carl Hansson) writes:
>
> I was sure 0! was undefined until I checked with my
> calculator and it said it was 1.
> 3! = 3*2*1 = 6
> 2! = 2*1 = 2
> 1! = 1
> 0! = 1
> -0! = -1
> -1! = -1
> -2! = -2
> -3! = -6
> according to my calculator (texas TI 82)
>
>
> Another question:
>
> Wats the factors in -2! ?
> -2*-1=2 according to me.
Warning!
Is this new or has it been going on for some time? That people learn
mathematics from calculators? This way, different calculators
represent different axiom systems. Maybe we get several mathematical
philosophy schools. Like: "The mathematical philosophy of TI 82" Or:
"The gospel according to HP 25".
Sorry, I couldn't resist.
--
Jon Haugsand
Dept. of Informatics, Univ. of Oslo, Norway, mailto:jonhaug@ifi.uio.no
http://www.ifi.uio.no/~jonhaug/, Pho/fax: +47-22852441/+47-22852401
Addr: Bredo Stabells v.15, N-0853 OSLO, NORWAY, Phone: +47-22952152
Subject: Re: Can this equation be solved for A?
From: Simon Read
Date: 15 Nov 96 16:42:39 GMT
"Bob Bridges" wrote:
>I have a problem in geometry which I've managed to reduce to
>
> H1 * H2
> C = -------
> H1 + H2
>
>...where Hn = SQRT(Ln^2 - A^2). Is it possible to isolate A and thus
>derive a general solution?
I can get you a polynomial out of the above, but beware, it's a big one!
Let's assume you mean H1 = sqrt(L1^2 - A^2) and H2 = sqrt(L2^2 - A^2)
so we've got
C(sqrt(L1^2 - A^2)+ sqrt(L2^2 - A^2)) =
sqrt(L1^2 - A^2).sqrt(L2^2 - A^2)
square everything (which introduces a few extra solutions - beware!)
C^2( (L1^2-A^2) + (L2^2-A^2) +2sqrt((L1^2-A^2)(L2^2-A^2)) )
= (L1^2-A^2)(L2^2-A^2)
Shuffle the square root onto the right hand side and square again
(yet more spurious solutions!):
[ (L1^2-A^2)(L2^2-A^2)/C^2 -(L1^2-A^2) -(L2^2-A^2) ]^2
= 4(L1^2-A^2)(L2^2-A^2)
so that gives you a polynomial, which can be solved by newton's method.
Then plug all solutions into the original equation and discard
three-quarters of them.
It might be easier just to use Newton's method on the original equation,
but that might have more trouble converging. Polynomials have some
well-defined methods which are pretty robust.
> Is there maybe a general solution for
>
> vA^4 + wA^2 + xA + y + zA^-1
>
That's an expression, not an equation, but I'm going to assume you're
trying to equate this to zero:
vA^4 + wA^2 + xA + y + zA^-1 = 0
then it's just a polynomial:
vA^5 + wA^3 + xA^2 + yA + z = 0
so if you know v, w, x, y, z you can solve for A.
Subject: Re: This Week's Finds in Mathematical Physics (Week 94)
From: singtech@teleport.com (Charles Cagle)
Date: 15 Nov 1996 10:54:31 -0800
In article <5688e2$43u@charity.ucr.edu>, baez@math.ucr.edu (John Baez) wrote:
[unneeded text snipped by relentless moderator - jb]
>First of all, remember that in quantum field theory, studying
>very small things is the same as studying things at very high
>energies. The reason is that in quantum mechanics you need to
>collide two particles at a large relative momentum p to make sure
Seems an odd way of putting it: 'relative momentum'. Isn't it a fact that
two particles actually have equal and opposite momenta with respect to
each other *even prior* to any possible interaction like an elastic
collision? If there were n particles in the universe wouldn't each
particle actually have, at all times, n-1 real momenta states? Can you
ever say that two particles with respect to one another actually have two
separate momenta the absolute value of either is not equal to the absolute
value of the other? So what do you mean by "a large relative momentum"
(as in the singular)?
--
C. Cagle
SingTech
Subject: Re: measure
From: David Ullrich
Date: Fri, 15 Nov 1996 11:06:04 -0600
JC wrote:
>
> David Ullrich wrote:
> >
> > JC wrote:
> > >
> > > JC wrote:
> > > >
> > > > Biblioteca matematica wrote:
> > > > >
> > > > > let C be Cantor set
> > > > > can you find a subset D of R that is omeomorph with C and such that its
> > > > > lebsgue measure isn't zero?
> > > >
> > > > No. Such a subset D would have to be compact (because C is) hence
> > > > closed in R, and nowhere dense (because it must be 0-dimensional,
> > > > hence interiorless). Such a set necessarily has 0 measure.
> > > >
> > > > JC
> > >
> > > Oops! Mea culpa. I was assuming that all copies of C in the reals
> > > were homeomorphic via a homeomorphism which extends to the reals.
> >
> > Huh? Actually I believe this is true, so assuming it should
> > not do any harm. How would this imply what you said about measure?
> > (Hint: It doesn't.)
> >
>
>
> Okay, then I'm really confused. The jist of my argument was this.
> Suppose you have copies of the Cantor set C1, C2 embedded in the
> reals. We can assume they both lie within [0,1], and (you claim,
> and I'm inclined to agree) there is a (monotone increasing)
> homeomorphism h:[0,1]->[0,1] taking C1 to C2. But such a map is
> uniformly continuous and therefore.... AH! The penny's dropped.
> I was assuming such a map preserved 0-measure for the simple
> reason that there is some K>0 s.t.
>
> |x-y| < K\epsilon => |hx-hy| < \epsilon
>
> which is of course bullsh*t.
Precisely. It's continuous, but not _that_ continuous.
Thinking about what I wrote the other day I hoped I hadn't said
that any homeomorphism from "the" Cantor set to "a" Cantor set
contained in R extends to a homeomorphism of R, but luckily I
didn't. I do believe that if C1 and C2 are as above you can
easily construct that monotone homeomorphism, bit by bit:
C1 and C2 are both totally disconnected, so you can
find a point p1 not in C1, such that there are points in C1
to the left of p1 and points in C1 to the right of p1. Now
p1 is contained in a maximal open interval (a1, b1) which
misses C1. Similarly construct (a2, b2) from C2. You map
(a1, b1) to (a2, b2) in the obvious way.
Now you do the same thing with the part of C1 to the
left of p1 and the part of C2 to the left of p2. Etc, and
you have a monotone bijection from the complement of C1 onto
the complement of C2. This extends to a continuous function
from R to R by monotonicity. (Well, it extends to a left-continuous
function by monotonicity, now the extension must be actually
continuous because there can't be any jumps, because the
Cantor sets do not contain any intervals. More or less...)
--
David Ullrich
?his ?s ?avid ?llrich's ?ig ?ile
(Someone undeleted it for me...)
Subject: Re: Autodynamics
From: "Todd K. Pedlar"
Date: Fri, 15 Nov 1996 12:08:21 -0600
Erik Max Francis wrote:
>
> Michael D. Painter wrote:
>
> > I'm more confused than ever now. If M = the solar mass then precession is
> > independent of the mass of the object.
> > This also implies that the orbits are circular? which they are not or
> > precession would not exist.
>
> Yes; in general relativity, more eccentric, smaller orbits have greater
> perihelion precession. In Autodynamics, supposedly, precession is
> independent of eccentricity; it only depends on the Sun's mass and the
> "orbital radius." Whether or not the orbits are assumed to be circular or
> whether or not "orbital radius" is another name for semimajor axis is
> questionable.
Certainly. This is one of the more damning evidences that their theory
of orbital precession is flawed. According to the formula which has
been
posted, a planet with earth's mass and mean orbital radius would precess
the same, no matter whether it had an orbit which is circular, or as
eccentric as pluto's. Doesn't this concern them, or have they even
thought
about it?
>
> However, general relativity can tell you Mercury's perihelion precession
> _from first principles_. You don't need to plug in the perihelion
> precession for another planet to "calibrate" the equation; general
> relativity just tells you what it is (and what it is for the other
> planets).
This is a fact which they conveniently disregard.
>
> Furthermore, Autodynamics' supposed predictions of the perihelion
> precession rates for the other planets are so high that I should think that
> these would have been observed by now. Have there been any dedicated
> efforts to measuring the precession rates for non-Mercury planets?
I don't believe that precession is seen for anyone but Mercury, at any
large level as they claim.
------------------------------------------------------------------
Todd K. Pedlar - Northwestern University - FNAL E835
Nuclear & Particle Physics Group
------------------------------------------------------------------
Phone: (847) 491-8630 (708) 840-8048 Fax: (847) 491-8627
------------------------------------------------------------------
WWW: http://numep1.phys.nwu.edu/tkp.html
------------------------------------------------------------------
Subject: Re: Fixed-point question
From: Larry Taylor
Date: 15 Nov 1996 16:54:52 GMT
Nick Halloway wrote:
>
>If a topological space X is compact and contractible, must it satisfy a
>fixed point theorem? That is, is there a compact contractible space
>X and continuous f: X --> X which has no fixed points?
>
>Please explain your terms, I'm a beginner at topology.
>
>
OK, I got this wrong last time, so I did some research. The following
is from the book _Fixed Point Theorems_ by D.R. Smart. Page 20.
The problem was open for 20 years until Kinoshita came up with the
following example in 1953. Fund. Math 40, 96-98.
In R^3, "take the union of a horizontal closed disc, a vertical cylinder
of unit height based on the edge of the disc, and a vertical sheet of
unit height and infinite length which spirals out from the axis of the
cylinder approaching closer and closer to the cylinder."
The mapping isn't described and is not immediately apparent to me. Nor
do I have access to the original reference. It would be easy to extend a
rotation of the outer cylinder to the vertical sheet, but how about the
base disk? It certainly had better not be mapped into itself. ???
Larry Taylor
Subject: Re: One combinatorial problem
From: Gareth McCaughan
Date: 15 Nov 1996 16:54:41 +0000
Alexander Burshteyn wrote:
> I wonder if anyone can offer any solution (or even a suggestion) about the
> following problem which came up in the course of my research:
>
> For every positive integer $n>1$ and every permutation $\tau \in S(n)$
> (i.e. of {1,...,n}, the following inequality holds:
>
> $
> \sum_{j=1}^{n} {
> \sum_{k=1}^{n} {
> \binomial{j+k-2,j-1} \times \binomial{2n-j-k,n-j} \times
> \binomial{ \tau (j) + \tau (k) - 2, \tau (j) - 1} \times
> \binomial{ 2n - \tau (j) - \tau (k), n - \tau (j)}
> }
> }
> > \binomial{2n-1,n} ^ 2
> $
>
> Even a proof or a pointer for the case $\tau = id(n)$ would be great.
Actually, it would be more than great: it would be enough, by
Chebyshev's[1] inequality (if (a[i]) are increasing then
sum of a[i].b[tau(i)] is maximal when tau is chosen so that
(b[i]) are increasing too).
I don't have a proof, though.
[1] If that happens not to be your preferred spelling, too bad.
--
Gareth McCaughan Dept. of Pure Mathematics & Mathematical Statistics,
gjm11@dpmms.cam.ac.uk Cambridge University, England.
Subject: Re: Laws of Flight
From: jaspevacek@mmm.com (John Spevacek)
Date: 15 Nov 1996 19:11:09 GMT
gerryq@indigo.ie (Gerry Quinn) wrote:
>I wonder can any aeronautical experts answer these questions:
>
>(i) If the density of air increases, does flight become easier, and if so in
>what proportions? Or does it depend on the velocity?
>
>(ii) Suppose a large dome was held at high pressure so as to create the
>maximum safe atmospheric density ( 10-20 kg/m3 I suppose ). Would
>human-powered flight be possible inside such a dome?
>
You can probably find a better newsgroup to post to, but here goes:
1) Bernoulli's equation is p + 1/2 (rho)*[(v*v) + (g*y)] = constant. This
must be applied along the streamlines above and below the wing. p is
pressure, rho is density, v is velocity, g is gravity and y is height.
For a wing, the velocity above the wing is lower than that below the
wing, so p is lower above the wing and thus it lifts (the g*y term is
insignificant). Similarly, if rho increases, the lift will also increase.
This is intuitively correct, since one cannot "fly" in space as there is
not air. And since the equation above is continuous, the behavior of
flight should follow a continuous curve.
2) Sure, human powered flight is already possible under normal conditions
(well, special normal conditions: low wind, low temperature, ...)
John
Subject: Problem with double integral
From: Adrian Burd
Date: Fri, 15 Nov 1996 14:07:25 -0600
I've been bashing my brain cells on the following problem without
success and was wondering if anyone out there might have any ideas.
The following integral arises in models of plant canopy structure. It
is basically the averaged leaf area projected normal to a beam of
light. In pseudo-tex
G = 1/(2*pi) int_0^{2*pi) d phi_L int_0^{pi/2} d theta_L
|cos(theta)*cos(theta_L) +
sin(theta)*sin(theta_L)*cos(phi-phi_L)|
where the normal to the leaf is the vector
n = (sin(theta_L)*cos(phi_L), sin(theta_L)*sin(phi_L), cos(phi_L))
and the beam direction is
m = (sin(theta)*cos(phi), sin(theta)*sin(phi), cos(phi))
In other words, G is just the integral over the upper hemisphere of
the absolute value of the dot product between the vectors m and n. So
phi is in [0,2*pi] and theta is in [0,pi/2] and similarly for phi_L
and theta_L. The angles are just the zenith angle and azimuth of the
beam and leaf normal respectively.
Without taking the absolute value, the integral is fine. However,
*with* the absolute value, one has to mess with the integration
limits.
One can show that |m.n| = ( exprn between |....| above) is non-negative
for (theta + theta_L) <= pi/2 so one can just go ahead and integrate
over theta_L between zero and (pi/2 - theta). This give one part of
the integral.
The next part, (theta + theta_L) > pi/2 has to be subdivided when
doing the phi_L integral in order to get all positive contributions.
Anyway, what happens when I do all this is that I get integrals that
appear to have no closed form solution. However, people publish
(without derivation) the value of the integral as being 1/2.
If anyone has any ideas on the evaluation of this I would be most
grateful.
Many thanks....
Adrian
--
_____________________________________________________________________
|
Adrian Burd, | Quidquid Latinae dictum sit,
Dpt. of Oceanography, | altum videtur.
Texas A&M; University, |
College Station , |---------------------------------------
Texas 77843 |
F: (409) 847 8879 | http://www-ocean.tamu.edu/~ecomodel
W: (409) 845 1115 |
_____________________________|_______________________________________
Disclaimer: I am not the official spokesperson for anyone, for which
organisations that use spokespersons are profoundly
grateful.
______________________________________________________________________
Subject: Re: Laws of Flight
From: Clark Dorman
Date: 15 Nov 1996 12:18:32 -0500
"Jack W. Crenshaw" writes:
> Gerry Quinn wrote:
> >
> > I wonder can any aeronautical experts answer these questions:
> >
> > (i) If the density of air increases, does flight become easier, and if so in
> > what proportions? Or does it depend on the velocity?
>
> Absolutely, it becomes easier. It also depends on the velocity, as
> usual. Lift and drag both depend on dynamic pressure, which is
>
> Q = (1/2) rho*v^2
>
> where rho is the density. Double the density, and you can lower the
> velocity. Pilots know this, and speak of density altitude -- the
> equivalent altitude at standard temperature (density goes down as temp
> goes up, of course).
It took me a while but I agree with your answer but not really your
reasoning. One of the important factors in flight is the lift to drag ratio.
If lift and drag both depend the same way on the density (i.e. linearly),
then an increase in density doesn't do any good.
However, the important factor here is the power: while lift and drag depend
on the velocity squared, the _power_ necessary for flight is a function of
the velocity cubed. By increasing the density, we decrease velocity by a
factor of the square root of the density; the power necessary for flight is
decreased by a factor of the density raised to the 3/2. The decrease in
power is what makes it "easier".
(Remember that power = energy / second. Energy = force * distance. Force is
in units (under mks) of kg * m /s^2. Energy is kg*m^2/s^2, power is
kg*m^2/s^3. So, it makes sense that power = kg*m/s^2 * m/s, or force times
velocity (V^2). Since force in this case is the drag, a function of V^2,
power is a function of V^3.)
> They know that takeoff speed and landing speed are
> higher, at higher density altitudes.
I presume that you mean that takeoff and landing speeds are lower at higher
densities and/or takeoff and landing speeds are higher at higher altitudes.
> > (ii) Suppose a large dome was held at high pressure so as to create the
> > maximum safe atmospheric density ( 10-20 kg/m3 I suppose ). Would
> > human-powered flight be possible inside such a dome?
>
> Good question, and the answer is, of course, yes, since human-powered
> flight is (just barely) possible without the dome. I think what you're
> suggesting is, wouldn't it be much _EASIER_ inside. Answer: yes.
> Especially if you also made it as cold as bearable.
Yes indeed, a lower temperature would allow a given density to occur at a
lower pressure. Assuming that your dome is pressure limited, a lower
temperature gives a higher density (universal gas law: PV = nRT).
Atmospheric density is normally 1.293 kg/m^3. Even if you could double that,
you would reduce the power necessary by 2.83. At 3 times atmospheric, you
would reduce the power by 5.2. On the other hand, don't underestimate the
difficulty in making a dome at higher than two or three atmospheres without
some serious leaking. If you build it though, please invite me for some
testing.
--
Clark Dorman "Evolution is cleverer than you are."
http://cns-web.bu.edu/pub/dorman/D.html -Francis Crick
Subject: Re: Rigorous statistical proof of Divine Authorship of the Bible?
From: Ariel Scolnicov
Date: 15 Nov 1996 20:43:50 +0200
rnh@gmrc.gecm.com (Richard Herring) writes:
> Herman Rubin (hrubin@b.stat.purdue.edu) wrote:
>
> >I agree almost completely with you. As to where it would be encoded,
> >there is the quite ancient belief, going back almost 2500 years, that
> >the first 5 books of the Hebrew Bible were given in that form directly
> >to Moses.
>
> Even the part recording the death of Moses?
According to the Midrash Agaddah ("interpretation according to
legends"), I think, Moses wrote those last verses and cried as he
wrote them. Note that, unlike certain people today, the people who
wrote this original legend were almost surely aware of what they were
doing, and probably didn't intend this to be interpretted as "gospel"
truth; they were just indicating awareness of the problem, and why it
shouldn't really bother anyone (I mean, if you accept divine
authorship *this* isn't going to stop you, and if you don't, surely
you have better reasons!).
A more serious anachronism is when verses are interpretted as Gimetria
(a way of coding numbers by letters). This was, uhmmm, stolen from the
Greeks (Hellenics, actually) 'round about the 2nd century BC. Gimetria
is usually the way numerological arguments about the bible
work. Again, if you believe in divine authorship this shouldn't bother
you, and if you don't -- you've probably got much better arguments.
