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Bill wrote:
Robert Bonomi wrote:
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Max wrote:
"Chris Friesen" wrote in message
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On 03/24/2010 11:18 AM, wrote:
Let's say you have a board that is 1" thick, 3" wide, 6" long.
The "ends" are where the grain shears off.
The "edges" are the thinner sides.
The "faces" are the wider sides.
But if I were to rip a thin strip off ...say 1/2" wide, so that the
smaller
"board" is 1 x 1/2 x 6...have the "edges" and "faces" changed places?
Is the "edge" still the thinner (1/2") side?
Personally I'd say yes. The edge should be the thinnest side that shows
long grain.
Of course it gets tricky if you have pieces with a square cross-section.
Incidentally, for cabinetmaking plywood the second dimension is the
grain direction. So an 8x4 sheet has the grain going the short way.
Chris
Lowe's has some 4X4 sheets.
What is one to do?
Do what a purported disconnected number recording at MIT once advised:
"The number you have called is imaginary,
Please rotate your phone 90 degrees, and try again."
Cute. You mean multiply it by exp^(i \PI /2).
No, I don't 'mean' that.
If I'd said that, it would *not* have been acurate reportage of the story
_as_I_heard_it_.
I've never seen the word
imaginary and degrees in the same sentence--no, never before.
Admittedly, it is a complex subject.
But, even with 'pure' imaginaries, it's still a matter of degree.
what's sqrt(-i), for example? "imaginary in the 2nd degree"? *GRIN*
sqrt(-i) = exp(-PI/4) or exp(3PI/4). Not "purely imaginary" at all--the
real part is +/- sqrt(2)/2. It's as plain as a point on the unit circle.
You could double-check that if a = sqrt(2)/2 - sqrt(2)/2*i, or b=-a,
then a^2=b^2=-i. Simple trig and De Moivre's Theorem...
I must be doing something wrong:
( sqrt(2)/2 - sqrt(2)/2*i ) **2
=(sqrt(2)/2)**2 -*((sqrt(2)/2*sqrt(2)/2*i) - (sqrt(2)/2*i)**2
= 2/4 -2*(2/4*i) - 2/4*(i*i)
= 2/4 - 4/4*i + 2/4
= 1-i
I'm tempted to make a facitous remark about an 'off by one' error.
Admittedly, I haven't played with this stuff for 30+ years, but I've got
a vague recollection of 'j' (the 'hyper-imaginary'??) as sqrt(-i).
Of course, on my first exposure to the imaginary exponential, I immediately
and rather vehemently questioned "exp(2*pi*i) = 1". As follows:
exp(0) = 1 by definition
exp(2*pi*i) = exp(0) 2 thins equal same thing, equal to each other
(2*pi*i) = 0 if bases (non-zero, and non-multiplicative-
identity) equal, exponents equal.
This rather upset the 8th grade math teacher. He *knew* my reasoning
was incorrect, but pointing out _where_ the flaw was was not obvious.
Simple reasoning on complex issues can lead one to into trouble. grin
Once it was clarified that '2*pi' was angular measure, things clarified
It is true that 2*pi == 0 (plus 1 revolution), even though neither of
the multiplicands is zero.