sorrpoint wrote: > In the meantime, people just keep talking in circles because there is no > real will to grapple with the reality of self-interest, [ .. ] basic greed [ .. ] Video61@tcq.net wrote: yep, the easy money crowd used easy money to line the pockets of the already filthy rich. the current free market crank science that got its start, after the last free market blowout, was a cleverly disguised as egalitarian, poverty reducing form of economics, the exact opposite of what it really was. it was a brilliant attempt at deception. the only problem is, its never worked, and it always ends the same, in depression. "once you get rid of the panderers to the wealthy out of the economics profession, then it can become a real science." "The only thing necessary for the triumph of evil is for good men to do nothing" . --Edmund Burke
ahahaha... Funny. I think it was Hawking's first wife who said that her job as Hawking's wife was to keep reminding him that he was not God. I'd say. It's a little hard to be God when you're a crumpled fool in a wheelchair making shit up that don't exist. The most annoying thing about physicists is not their crackpottery but their insufferable pomposity.
Uhm, did I say that 0 = 1 has not a sensible model in physics? Take a
look at the following picture (but first take a non-proportional font):
plus + ___________
O contact
:::------------=====-------------O 0
| spring ______ _ |
| coil //// | / \ 1
| \\\\ | | o | bell
|____________//// | \ _ /
minus -
This apparatus is called a "buzzer" in English, if I'm informed well.
It's a two state machine. Let's call the states 1 and 0. If the machine
is in state 0, an electric current will flow through the coil and will
make it magnetic. So the contact will be broken and the bell rings: 1.
But if the contact is broken, the string will pull back the contact: 0.
And so on, and so forth. So 1 causes 0 and 0 causes 1. Like a paradox,
which is repeatedly resolved in time. Let x be a boolean with values 1
and 0, then the process is described by: x := ~ x . Sloppy: x =/= x .
<NL>
http://nl.wikipedia.org/wiki/Kegel_(ruimtelijk_figuur)
Inhoud van een kegel = Pi.r^2.h/3
Bovenrand schaal straal = R , onderrand schaal straal = r ,
hoogte schaal = h .
De schaal is een verschil van twee kegels.
Grote kegel = Pi.R^2.H/3
Kleine kegel = Pi.r^2.(H-h)/3
Met (H-h)/H = r/R
dus (H-h).R = H.r => H.(R-r) = R.h , H = h.R/(R-r)
(H-h) = h.R/(R-r).r/R = h.r/(R-r)
Grote kegel = Pi.R^2.H/3 = Pi.R^2.h.R/(R-r)/3
Kleine kegel = Pi.r^2.(H-h)/3 = Pi.r^2.h.r/(R-r)/3
Totale inhoud schaal = Pi.R^2.h.R/(R-r)/3 - Pi.r^2.h.r/(R-r)/3
= Pi.(R^3 - r^3).(R-r).h/3
= Pi.h/3.(R^2 + R.r + r^2)
Inderdaad: http://www.wisfaq.nl/show3archive.asp?id=25289&j=2004
Nu het vullen van de schaal op een afstand x van de bovenrand.
Dat zijn twee afgeknotte kegels die dezelfde inhoud moeten hebben.
Straal van de appelmoes tot waar gevuld is een lineaire interpolatie:
m = R - (R-r)/h.x . Inderdaad x = 0 dan m = R , x = h dan m = r .
Kegel 1 = Pi.x/3.[ R^2 + R.(R - (R-r)/h.x) + (R - (R-r)/h.x)^2 ] =
Kegel 2 = Pi.(h-x)/3.[ r^2 + r.(R - (R-r)/h.x) + (R - (R-r)/h.x)^2 ]
MAPLE geeft als reele oplossing:
x = h/(R-r).(R + (- 4.R^3 - 4.r^3)^(1/3) / 2 )
= h/(R-r).(R - (R^3/2 + r^3/2)^(1/3) )
De grootte van de straal ter plaatse is (R^3/2 + r^3/2)^(1/3) = m .
Het is dus inderdaad niet zo simpel!
Met vriendelijke groet:
Han de Bruijn
N.B. Tweedimensionaal is op analoge wijze: m = (R^2/2 + r^2/2)^(1/2)
</NL>
> A := Pi*x/3*(R^2 + R*(R-(R-r)*x/h) + (R-(R-r)/h*x)^2);
/ 2 / (R - r) x\ / (R - r) x\2\
Pi x |R + R |R - ---------| + |R - ---------| |
\ \ h / \ h / /
A := ------------------------------------------------
3
> B := Pi*(h-x)/3*(r^2 + r*(R - (R-r)/h*x) + (R - (R-r)/h*x)^2 );
/ 2 / (R - r) x\ / (R - r) x\2\
Pi (h - x) |r + r |R - ---------| + |R - ---------| |
\ \ h / \ h / /
B := ------------------------------------------------------
3
> solve(A-B=0,x);
/ 3 3 1/3\
| (-4 R - 4 r ) |
h |R + -----------------|
\ 2 /
-------------------------,
R - r
/ 3 3 1/3 \
| (-4 R - 4 r ) 1/2 3 3 (1/3)|
h |R - ----------------- - 1/4 I 3 (-4 R - 4 r ) |
\ 4 /
----------------------------------------------------------,
R - r
/ 3 3 1/3 \
| (-4 R - 4 r ) 1/2 3 3 (1/3)|
h |R - ----------------- + 1/4 I 3 (-4 R - 4 r ) |
\ 4 /
----------------------------------------------------------
R - r
Re: Math Mag article on volume of liquid in tilted frustum of cone
dmr5713@gmail.com wrote:
> A surprisingly complicated problem is the following. Consider a
> container shaped like a frustum of a right circular cone (e.g. a
> styrofoam coffee cup or a plastic yogurt container), with radius of
> the bottom r, radius of the top R, and perpendicular height h. Fill
> it with liquid and tilt it, pouring out enough liquid so that the
> highest point of the bottom of the container is level with the lowest
> point of the top rim of the container. That is, the surface of the
> liquid should now just touch the highest point of the edge of the
> bottom and the lowest point of the rim. Ignoring surface effects,
> what's the volume of the liquid in the container in terms of r, R & h?
>
> There was an article in Mathematics Magazine (or perhaps the American
> Mathematical Monthly, but I'm pretty sure it was Mathematics Magazine)
> some time in the last couple of decades that gave a neat way of
> solving this and some similar problems, but I can't seem to locate
> it. Does anyone have the reference?
I can't see how to solve this without calculus (i.e. integration).