Regarding that business up above about dividing a circle into 3 equal
parts using a square, blah blah blah ... maybe I don't remember my
geometry so well. Checked back in that table I mentioned in "Proven Shop
Tips" for dividing a circle into n equal parts, and sure enough, the
number to multiply the diameter of a circle by to divide into 6 equal
parts is ... exactly 0.5.

So anyone got the proof handy that a hexagon with sides of length s can
be inscribed in a circle whose radius equals s? I have my old algebra
and calculus books, but no geometry.


--
Found--the gene that causes belief in genetic determinism

On Tue, 19 May 2009 18:07:58 -0700, David Nebenzahl
wrote:

....
So anyone got the proof handy that a hexagon with sides of length s can
be inscribed in a circle whose radius equals s? I have my old algebra
and calculus books, but no geometry.

It's easy enough to show using Trigonometry, but that doesn't
constitute a geometric proof, which as I recall has to be done with
only compass and straight edge. But it's simple enough to demonstrate
with a compass and straight edge. I don't know whether demonstration
by construction constitutes a formal geometric proof, or not.

Set your compass to a convenient radius and draw a circle. Without
changing the compass setting strike an arc from any point on the
circle that intersects the circle. From that intersection, strike
another intersecting arc. Continue around the circle and, if done
carefully enough, the 6th arc will pass through the original point.
Since all arcs have the same radius, all the chords connecting the
intersections are the same length and equal to the radius of the arc
which is also the radius of the circle. Connect each point of
intersection with its neighbors using a straight line. By definition,
6 sides, all of the same length, constitute a regular hexagon.

Tom Veatch
Wichita, KS
USA

An armed society is a polite society.
Manners are good when one may have to back up his acts with his life.
Robert A. Heinlein

On Tue, 19 May 2009 20:47:59 -0500, Tom Veatch wrote:

. I don't know whether demonstration
by construction constitutes a formal geometric proof, or not.


Given your demonstration, I believe you can safely append:


QED







Regards,

Tom Watson
http://home.comcast.net/~tjwatson1/

"David Nebenzahl" wrote:

So anyone got the proof handy that a hexagon with sides of length s
can be inscribed in a circle whose radius equals s? I have my old
algebra and calculus books, but no geometry.

Remember the first three (3) plane geometry proofs?

1) Side-Angle-Side
2) Side-Side-Side
3) Angle-Side-Angle

Side-Side-Side works for me.

Lew

David Nebenzahl wrote:

Regarding that business up above about dividing a circle into 3 equal
parts using a square, blah blah blah ... maybe I don't remember my
geometry so well. Checked back in that table I mentioned in "Proven Shop
Tips" for dividing a circle into n equal parts, and sure enough, the
number to multiply the diameter of a circle by to divide into 6 equal
parts is ... exactly 0.5.

So anyone got the proof handy that a hexagon with sides of length s can
be inscribed in a circle whose radius equals s? I have my old algebra
and calculus books, but no geometry.

An equilateral triangle has equal sides and three 60-degree angles.
Arrange six of them with sides s in an array with one common vertex,
and you will have the inscribed hexagon.
--
Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently.

On 5/19/2009 6:47 PM Tom Veatch spake thus:

On Tue, 19 May 2009 18:07:58 -0700, David Nebenzahl
wrote:

...
So anyone got the proof handy that a hexagon with sides of length s can
be inscribed in a circle whose radius equals s? I have my old algebra
and calculus books, but no geometry.

Set your compass to a convenient radius and draw a circle. Without
changing the compass setting strike an arc from any point on the
circle that intersects the circle. From that intersection, strike
another intersecting arc. Continue around the circle ...

That's a demonstration, not a proof. But you knew that.

I'm interested in the proof. It can't be all that complicated.


--
Found--the gene that causes belief in genetic determinism

On 5/19/2009 7:02 PM Lew Hodgett spake thus:

"David Nebenzahl" wrote:

So anyone got the proof handy that a hexagon with sides of length s
can be inscribed in a circle whose radius equals s? I have my old
algebra and calculus books, but no geometry.

Remember the first three (3) plane geometry proofs?

1) Side-Angle-Side
2) Side-Side-Side
3) Angle-Side-Angle

Side-Side-Side works for me.

Expand, please. Don't know exactly how this proof works.


--
Found--the gene that causes belief in genetic determinism

"David Nebenzahl" wrote in message
.com...
On 5/19/2009 6:47 PM Tom Veatch spake thus:

On Tue, 19 May 2009 18:07:58 -0700, David Nebenzahl
wrote:

...
So anyone got the proof handy that a hexagon with sides of length s can
be inscribed in a circle whose radius equals s? I have my old algebra and
calculus books, but no geometry.

Set your compass to a convenient radius and draw a circle. Without
changing the compass setting strike an arc from any point on the
circle that intersects the circle. From that intersection, strike
another intersecting arc. Continue around the circle ...

That's a demonstration, not a proof. But you knew that.

I'm interested in the proof. It can't be all that complicated.

It is well-known that one cannot trisect an angle with a straight-edge and
compass, so I don't think you'll get a proof with that approach. On the
other hand, using division you can divide 360 degrees, or 2*Pi radians by 6
to get the angle for each slice of the pie The rest has already been
discussed (side-side-side).

Bill

"David Nebenzahl" wrote:

Expand, please. Don't know exactly how this proof works.

Plug " Side-Side-Side" into Google, should keep you out of trouble for
a couple of hours, especially the congruent triangle proofs.

Lew

alexy wrote:

David Nebenzahl wrote:

Regarding that business up above about dividing a circle into 3 equal
parts using a square, blah blah blah ... maybe I don't remember my
geometry so well. Checked back in that table I mentioned in "Proven Shop
Tips" for dividing a circle into n equal parts, and sure enough, the
number to multiply the diameter of a circle by to divide into 6 equal
parts is ... exactly 0.5.

So anyone got the proof handy that a hexagon with sides of length s can
be inscribed in a circle whose radius equals s? I have my old algebra
and calculus books, but no geometry.

An equilateral triangle has equal sides and three 60-degree angles.
Arrange six of them with sides s in an array with one common vertex,
and you will have the inscribed hexagon.

Or, put another way:

Consider a regular hexagon. Draw line segments from the center of the
hexagon to each of the six vertices. These six equal angles at the
center must add up to 360, so each is 60. Since the triangles are
isosceles, and their angles add to 180, they are also equilateral. So
the side of the hexagon is equal to the length of the line form the
center to a vertex on the hexagon, which is the radius of the circle.
--
Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently.

On 5/19/2009 7:18 PM alexy spake thus:

David Nebenzahl wrote:

Regarding that business up above about dividing a circle into 3 equal
parts using a square, blah blah blah ... maybe I don't remember my
geometry so well. Checked back in that table I mentioned in "Proven Shop
Tips" for dividing a circle into n equal parts, and sure enough, the
number to multiply the diameter of a circle by to divide into 6 equal
parts is ... exactly 0.5.

So anyone got the proof handy that a hexagon with sides of length s can
be inscribed in a circle whose radius equals s? I have my old algebra
and calculus books, but no geometry.

An equilateral triangle has equal sides and three 60-degree angles.
Arrange six of them with sides s in an array with one common vertex,
and you will have the inscribed hexagon.

I don't see any proof in there, only an assertion.


--
Found--the gene that causes belief in genetic determinism

On 5/19/2009 7:42 PM alexy spake thus:

alexy wrote:

David Nebenzahl wrote:

Regarding that business up above about dividing a circle into 3 equal
parts using a square, blah blah blah ... maybe I don't remember my
geometry so well. Checked back in that table I mentioned in "Proven Shop
Tips" for dividing a circle into n equal parts, and sure enough, the
number to multiply the diameter of a circle by to divide into 6 equal
parts is ... exactly 0.5.

So anyone got the proof handy that a hexagon with sides of length s can
be inscribed in a circle whose radius equals s? I have my old algebra
and calculus books, but no geometry.

An equilateral triangle has equal sides and three 60-degree angles.
Arrange six of them with sides s in an array with one common vertex,
and you will have the inscribed hexagon.

Or, put another way:

Consider a regular hexagon. Draw line segments from the center of the
hexagon to each of the six vertices. These six equal angles at the
center must add up to 360, so each is 60. Since the triangles are
isosceles, and their angles add to 180, they are also equilateral. So
the side of the hexagon is equal to the length of the line form the
center to a vertex on the hexagon, which is the radius of the circle.

That sounds better. (Don't know if it constitutes a rigorous proof or
not, but it satisfies my "itching".)


--
Found--the gene that causes belief in genetic determinism

David Nebenzahl wrote:

On 5/19/2009 7:18 PM alexy spake thus:

David Nebenzahl wrote:

Regarding that business up above about dividing a circle into 3 equal
parts using a square, blah blah blah ... maybe I don't remember my
geometry so well. Checked back in that table I mentioned in "Proven Shop
Tips" for dividing a circle into n equal parts, and sure enough, the
number to multiply the diameter of a circle by to divide into 6 equal
parts is ... exactly 0.5.

So anyone got the proof handy that a hexagon with sides of length s can
be inscribed in a circle whose radius equals s? I have my old algebra
and calculus books, but no geometry.

An equilateral triangle has equal sides and three 60-degree angles.
Arrange six of them with sides s in an array with one common vertex,
and you will have the inscribed hexagon.

I don't see any proof in there, only an assertion.

Look deeper. This is a proof of the "from which it can be clearly
seen..." type that occasionally drove me batty!

--
Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently.

"David Nebenzahl" wrote:

Expand, please. Don't know exactly how this proof works.
=================================
This is another purely graphical soultion.

http://mathworld.wolfram.com/Hexagon.html

Lew









Plug " Side-Side-Side" into Google, should keep you out of trouble
for a couple of hours, especially the congruent triangle proofs.

Lew

This is another purely graphical soultion.

http://mathworld.wolfram.com/Hexagon.html

Lew


The illustration with the A, B, C, D and E circles gives the OP his 3
pie pieces, if he just puts the triangle inside the hexagon.


--

-MIKE-

"Playing is not something I do at night, it's my function in life"
--Elvin Jones (1927-2004)
--
http://mikedrums.com

---remove "DOT" ^^^^ to reply

In article , alexy wrote:

Consider a regular hexagon. Draw line segments from the center of the
hexagon to each of the six vertices. These six equal angles at the
center must add up to 360, so each is 60.

OK so far...

Since the triangles are
isosceles, and their angles add to 180, they are also equilateral.

... but you just went astray there.

That's not sufficient to prove that the triangles are equilateral, since the
angles add to 180 in *all* triangles.

This may be what you meant to say:

Since the angle at the vertex of each triangle is 60 degrees, the sum of the
angles at the base is 180 - 60 = 120 degrees. Since each triangle is
isosceles, the angles at the base are equal, and (since they add to 120)
therefore also 60 degrees. The triangles are therefore equiangular, and
therefore equilateral.

So
the side of the hexagon is equal to the length of the line form the
center to a vertex on the hexagon, which is the radius of the circle.

In article , "Bill" wrote:

"David Nebenzahl" wrote in message
s.com...
On 5/19/2009 6:47 PM Tom Veatch spake thus:

On Tue, 19 May 2009 18:07:58 -0700, David Nebenzahl
wrote:

...
So anyone got the proof handy that a hexagon with sides of length s can
be inscribed in a circle whose radius equals s? I have my old algebra and
calculus books, but no geometry.

Set your compass to a convenient radius and draw a circle. Without
changing the compass setting strike an arc from any point on the
circle that intersects the circle. From that intersection, strike
another intersecting arc. Continue around the circle ...

That's a demonstration, not a proof. But you knew that.

I'm interested in the proof. It can't be all that complicated.

It is well-known that one cannot trisect an angle with a straight-edge and
compass, so I don't think you'll get a proof with that approach. On the
other hand, using division you can divide 360 degrees, or 2*Pi radians by 6
to get the angle for each slice of the pie The rest has already been
discussed (side-side-side).

Nobody's attempting to trisect an angle in that approach; in fact, it's
essentially the same method as Euclid's proof, a link to which was already
posted up-thread.

(Doug Miller) wrote:

In article , alexy wrote:

Consider a regular hexagon. Draw line segments from the center of the
hexagon to each of the six vertices. These six equal angles at the
center must add up to 360, so each is 60.

OK so far...

Since the triangles are
isosceles, and their angles add to 180, they are also equilateral.

.. but you just went astray there.

That's not sufficient to prove that the triangles are equilateral, since the
angles add to 180 in *all* triangles.
But I also pointed out that the triangle was isosceles. Isosceles and
one 60 degree angle are necessary and sufficient conditions for an
equilateral triangle (in Euclidean geometry).

This may be what you meant to say:

Since the angle at the vertex of each triangle is 60 degrees, the sum of the
angles at the base is 180 - 60 = 120 degrees. Since each triangle is
isosceles, the angles at the base are equal, and (since they add to 120)
therefore also 60 degrees. The triangles are therefore equiangular, and
therefore equilateral.

That is what I said. You just have to read between the lines g.

So
the side of the hexagon is equal to the length of the line form the
center to a vertex on the hexagon, which is the radius of the circle.

--
Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently.

On Tue, 19 May 2009 18:07:58 -0700, David Nebenzahl
wrote:

Regarding that business up above about dividing a circle into 3 equal
parts using a square, blah blah blah ... maybe I don't remember my
geometry so well. Checked back in that table I mentioned in "Proven Shop
Tips" for dividing a circle into n equal parts, and sure enough, the
number to multiply the diameter of a circle by to divide into 6 equal
parts is ... exactly 0.5.

So anyone got the proof handy that a hexagon with sides of length s can
be inscribed in a circle whose radius equals s? I have my old algebra
and calculus books, but no geometry.


Well, I havn't seen the proof but dividing a circle into thirds is
very easy with a compass. Set your radius, and keep it there. Draw
the circle, set the compass point on the circle and draw an arc inside
the circle. Repeat 5 times using an intersect as another pivot point.
You will get a perfect 6-petal flower.

1. Take the circle, and draw a radius.
2. Use a compass to measure from the intersection of the radius and
circle, to the center.
3. Scribe a circle from that point.
4. connect the center to the 2 new intersections
5. You now have 2 equilateral triangles inside the circle (all 3 sides
are equal. - they're radii)
you now have a third of a circle (or 2 sixths)
6. continue all the way around for the hexagon

shelly

On 5/20/2009 7:20 PM spake thus:

1. Take the circle, and draw a radius.
2. Use a compass to measure from the intersection of the radius and
circle, to the center.
3. Scribe a circle from that point.
4. connect the center to the 2 new intersections
5. You now have 2 equilateral triangles inside the circle (all 3 sides
are equal. - they're radii)
you now have a third of a circle (or 2 sixths)
6. continue all the way around for the hexagon

Like more than half the respondents to this thread, you completely
missed the point.

I know how to do that. I wasn't asking for a demonstration; I was asking
for a *proof*.

(Even though your description contains some of the elements of a proof.)


--
Found--the gene that causes belief in genetic determinism

"David Nebenzahl" wrote in message
.com...
On 5/20/2009 7:20 PM spake thus:

1. Take the circle, and draw a radius.
2. Use a compass to measure from the intersection of the radius and
circle, to the center.
3. Scribe a circle from that point.
4. connect the center to the 2 new intersections
5. You now have 2 equilateral triangles inside the circle (all 3 sides
are equal. - they're radii)
you now have a third of a circle (or 2 sixths)
6. continue all the way around for the hexagon

Like more than half the respondents to this thread, you completely missed
the point.

I know how to do that. I wasn't asking for a demonstration; I was asking
for a *proof*.

State the axioms you wish to start with, and we'll take it from there. I'm
sure someone here knows how to write a proof. A calculation based
explanation is a good proof if one axiomizes high school geometry. This
seems like the right approach here in rec.woodworking!

"David Nebenzahl" wrote in message
.com...
On 5/20/2009 7:20 PM spake thus:

1. Take the circle, and draw a radius.
2. Use a compass to measure from the intersection of the radius and
circle, to the center.
3. Scribe a circle from that point.
4. connect the center to the 2 new intersections
5. You now have 2 equilateral triangles inside the circle (all 3
sides
are equal. - they're radii)
you now have a third of a circle (or 2 sixths)
6. continue all the way around for the hexagon

Like more than half the respondents to this thread, you completely
missed the point.

I know how to do that. I wasn't asking for a demonstration; I was
asking
for a *proof*.

(Even though your description contains some of the elements of a
proof.)


See:
http://www.nvcc.edu/home/tstreilein/.../inscribe4.htm

for the proof.

Len

On 5/21/2009 9:25 AM Len spake thus:

"David Nebenzahl" wrote in message
.com...

Like more than half the respondents to this thread, you completely
missed the point.

I know how to do that. I wasn't asking for a demonstration; I was
asking for a *proof*.

(Even though your description contains some of the elements of a
proof.)

See:
http://www.nvcc.edu/home/tstreilein/.../inscribe4.htm

for the proof.

Thank you. That was exactly what I was looking for.

There; was that so hard?


--
Found--the gene that causes belief in genetic determinism

"David Nebenzahl" wrote:

Thank you. That was exactly what I was looking for.

Same proof I gave you almost a week ago.

Lew

"David Nebenzahl" wrote in message
.com...
On 5/21/2009 9:25 AM Len spake thus:

"David Nebenzahl" wrote in message
.com...

Like more than half the respondents to this thread, you completely
missed the point.

I know how to do that. I wasn't asking for a demonstration; I was
asking for a *proof*.

(Even though your description contains some of the elements of a
proof.)

See:
http://www.nvcc.edu/home/tstreilein/.../inscribe4.htm

for the proof.


It seems to be based on calculation (360/6), but obscures that fact; I would
expect better work from a math major.

"Bill" wrote:


"David Nebenzahl" wrote in message
s.com...
On 5/21/2009 9:25 AM Len spake thus:

"David Nebenzahl" wrote in message
.com...

Like more than half the respondents to this thread, you completely
missed the point.

I know how to do that. I wasn't asking for a demonstration; I was
asking for a *proof*.

(Even though your description contains some of the elements of a
proof.)

See:
http://www.nvcc.edu/home/tstreilein/.../inscribe4.htm

for the proof.


It seems to be based on calculation (360/6), but obscures that fact; I would
expect better work from a math major.

LOL!

Obviously, as your previous post hinted, the OP didn't really want a
formal proof. (He may not realize that is not what he wanted, but the
fact that this question was raised because of not having an old
geometry text is a pretty good clue.) What he wanted was a logical
demonstration based on facts he accepted, with steps he didn't have to
figure out. Note the range of logically identical responses here that
have been dismissed as "mere demonstrations" or accepted as "proofs"
depending on the number and detail of the steps explicitly stated, and
whether the steps were numbered and labeled "proof" g.

--
Alex -- Replace "nospam" with "mail" to reply by email. Checked infrequently.

Obviously, as your previous post hinted, the OP didn't really want a
formal proof. (He may not realize that is not what he wanted, but the
fact that this question was raised because of not having an old
geometry text is a pretty good clue.) What he wanted was a logical
demonstration based on facts he accepted, with steps he didn't have to
figure out. Note the range of logically identical responses here that
have been dismissed as "mere demonstrations" or accepted as "proofs"
depending on the number and detail of the steps explicitly stated, and
whether the steps were numbered and labeled "proof" g.


Reminds me of a quote from a physics professor.

"You want an easy proof for the law of gravity? Step out of the window."


--

-MIKE-

"Playing is not something I do at night, it's my function in life"
--Elvin Jones (1927-2004)
--
http://mikedrums.com

---remove "DOT" ^^^^ to reply

-MIKE- wrote:

Reminds me of a quote from a physics professor.

"You want an easy proof for the law of gravity? Step out of the window."

Obviously, that proof makes the assumption that the classroom in
question is in an inertial reference frame.

Chris

Chris Friesen wrote:
-MIKE- wrote:

Reminds me of a quote from a physics professor.

"You want an easy proof for the law of gravity? Step out of the window."

Obviously, that proof makes the assumption that the classroom in
question is in an inertial reference frame.

Chris

The window was in a big frame, on the 5th floor. :-)


--

-MIKE-

"Playing is not something I do at night, it's my function in life"
--Elvin Jones (1927-2004)
--
http://mikedrums.com

---remove "DOT" ^^^^ to reply

"alexy" wrote in message
...
"Bill" wrote:


"David Nebenzahl" wrote in message
rs.com...
On 5/21/2009 9:25 AM Len spake thus:

"David Nebenzahl" wrote in message
.com...

Like more than half the respondents to this thread, you completely
missed the point.

I know how to do that. I wasn't asking for a demonstration; I was
asking for a *proof*.

(Even though your description contains some of the elements of a
proof.)

See:
http://www.nvcc.edu/home/tstreilein/.../inscribe4.htm

for the proof.


It seems to be based on calculation (360/6), but obscures that fact; I
would
expect better work from a math major.

LOL!

Obviously, as your previous post hinted, the OP didn't really want a
formal proof. (He may not realize that is not what he wanted, but the
fact that this question was raised because of not having an old
geometry text is a pretty good clue.) What he wanted was a logical
demonstration based on facts he accepted, with steps he didn't have to
figure out. Note the range of logically identical responses here that
have been dismissed as "mere demonstrations" or accepted as "proofs"
depending on the number and detail of the steps explicitly stated, and
whether the steps were numbered and labeled "proof" g.

--
Alex -- Replace "nospam" with "mail" to reply by email. Checked
infrequently.

Your LOL is well taken. I'm not sure whether one needs the numbers in
between the fractions (like sqrt(2)) for woodworking, nor any negative
numbers, imaginary numbers, non-real complex numbers, nor probably any
numbers bigger than 500. Maybe that's why those aren't marked on the ruler.


Bill

Bill wrote:

Your LOL is well taken. I'm not sure whether one needs the numbers in
between the fractions (like sqrt(2)) for woodworking, nor any negative
numbers, imaginary numbers, non-real complex numbers, nor probably any
numbers bigger than 500. Maybe that's why those aren't marked on the ruler.

sqrt(2) is useful to find the length of the long side of a 45/45/90
triangle. Similarly, 1/2/sqrt(3) are the sides of a 30/60/90 triangle.

Numbers bigger than 500 are useful when working in millimetres.

I'm up in Canada and I know a guy who does everything in mm. Although
the initial conversion of regular North American lumber dimensions to mm
is a bit of a pain, it makes subsequent math a lot simpler. And of
course all the Euro stuff just works...

Chris

"Chris Friesen" wrote in message
el...
Bill wrote:

Your LOL is well taken. I'm not sure whether one needs the numbers in
between the fractions (like sqrt(2)) for woodworking, nor any negative
numbers, imaginary numbers, non-real complex numbers, nor probably any
numbers bigger than 500. Maybe that's why those aren't marked on the
ruler.

sqrt(2) is useful to find the length of the long side of a 45/45/90
triangle.

Do you think 1 53/128 would suffice (somebody with good eyesight might be
able to mark it off a ruler with 64th's). I'd do better with a micrometer.
I hope the wood is very stable.





Similarly, 1/2/sqrt(3) are the sides of a 30/60/90 triangle.

Numbers bigger than 500 are useful when working in millimetres.

I'm up in Canada and I know a guy who does everything in mm. Although
the initial conversion of regular North American lumber dimensions to mm
is a bit of a pain, it makes subsequent math a lot simpler. And of
course all the Euro stuff just works...

Chris

Bill wrote:

Your LOL is well taken. I'm not sure whether one needs the numbers in
between the fractions (like sqrt(2)) for woodworking, nor any negative
numbers, imaginary numbers, non-real complex numbers, nor probably any
numbers bigger than 500. Maybe that's why those aren't marked on the ruler.


It might depend on what you're doing. The ribs at

http://www.iedu.com/DeSoto/Projects/Stirling/Heat.html

needed to be cut so that the length along the parabola was exactly four
feet (the mirror width) and with accuracy to provide a good optical
focus along the entire eight-foot length - and...

....the tenoned parts shown at the bottom of

http://www.iedu.com/DeSoto/Projects/Bevel/

were for silverware trays with diagonal dividers; these were the divider
blanks, and they needed to /exactly/ fit (on /both/ ends ).

And no, none of the numbers needed were marked on any of my rulers.

--
Morris Dovey
DeSoto Solar
DeSoto, Iowa USA
http://www.iedu.com/DeSoto/

"Morris Dovey" wrote in message
...
It might depend on what you're doing. The ribs at

http://www.iedu.com/DeSoto/Projects/Stirling/Heat.html

needed to be cut so that the length along the parabola was exactly four
feet (the mirror width) and with accuracy to provide a good optical focus
along the entire eight-foot length - and...

...the tenoned parts shown at the bottom of

http://www.iedu.com/DeSoto/Projects/Bevel/

were for silverware trays with diagonal dividers; these were the divider
blanks, and they needed to /exactly/ fit (on /both/ ends ).

And no, none of the numbers needed were marked on any of my rulers.

--
Morris Dovey



Interesting projects! My wife is supportive of amost any outlay for tools
as long as I build her some "bird-related" stuff (feeders, houses, etc).
Birds
don't tend to be particular beyond a 16th of an inch.

Bill

Morris Dovey wrote:
And no, none of the numbers needed were marked on any of my rulers.


Am I the only one would rather mark than measure?

Like if I have a piece of trim that needs to fit between A and B,
I don't measure A to B then measure that out on the trim.
I hold up the trim between A and B and mark the trim.


--

-MIKE-

"Playing is not something I do at night, it's my function in life"
--Elvin Jones (1927-2004)
--
http://mikedrums.com

---remove "DOT" ^^^^ to reply

-MIKE- wrote:
Morris Dovey wrote:
And no, none of the numbers needed were marked on any of my rulers.


Am I the only one would rather mark than measure?

Like if I have a piece of trim that needs to fit between A and B,
I don't measure A to B then measure that out on the trim.
I hold up the trim between A and B and mark the trim.


Before some smarta$$ says, what do you do with a 12' piece of crown
molding....
I obviously call a couple friends to come over and hold it in place for
me. duh.


--

-MIKE-

"Playing is not something I do at night, it's my function in life"
--Elvin Jones (1927-2004)
--
http://mikedrums.com

---remove "DOT" ^^^^ to reply

-MIKE- wrote:
-MIKE- wrote:
Morris Dovey wrote:
And no, none of the numbers needed were marked on any of my rulers.

Am I the only one would rather mark than measure?

Like if I have a piece of trim that needs to fit between A and B,
I don't measure A to B then measure that out on the trim.
I hold up the trim between A and B and mark the trim.


Before some smarta$$ says, what do you do with a 12' piece of crown
molding....
I obviously call a couple friends to come over and hold it in place for
me. duh.

What do you do with a 12' piece of crown molding on an outside corner
that isn't square?

Chris

-MIKE- wrote:
Morris Dovey wrote:
And no, none of the numbers needed were marked on any of my rulers.

Am I the only one would rather mark than measure?

Like if I have a piece of trim that needs to fit between A and B,
I don't measure A to B then measure that out on the trim.
I hold up the trim between A and B and mark the trim.

If it'll help you feel better, I neither marked /nor/ measured for those
projects - everything was cut from unmarked stock and then assembled as cut.

I did have drawings for the silverware trays because the customer needed
something to sign off on, but the drawings for the parabolic trough came
along after the fact, to document what had been done.

For stuff I don't need to be fussy about, I've had to switch to a light
touch with a knife - my eyes just aren't good enough any longer to split
a pencil mark...

--
Morris Dovey
DeSoto Solar
DeSoto, Iowa USA
http://www.iedu.com/DeSoto/

"Bill" wrote in message
...

"alexy" wrote in message
...
"Bill" wrote:


"David Nebenzahl" wrote in message
rs.com...
On 5/21/2009 9:25 AM Len spake thus:

"David Nebenzahl" wrote in message
.com...

Like more than half the respondents to this thread, you
completely
missed the point.

I know how to do that. I wasn't asking for a demonstration; I
was
asking for a *proof*.

(Even though your description contains some of the elements
of a
proof.)

See:

http://www.nvcc.edu/home/tstreilein/.../inscribe4.htm

for the proof.


It seems to be based on calculation (360/6), but obscures that
fact; I
would
expect better work from a math major.

LOL!

Obviously, as your previous post hinted, the OP didn't really
want a
formal proof. (He may not realize that is not what he wanted,
but the
fact that this question was raised because of not having an old
geometry text is a pretty good clue.) What he wanted was a
logical
demonstration based on facts he accepted, with steps he didn't
have to
figure out. Note the range of logically identical responses here
that
have been dismissed as "mere demonstrations" or accepted as
"proofs"
depending on the number and detail of the steps explicitly
stated, and
whether the steps were numbered and labeled "proof" g.

--
Alex -- Replace "nospam" with "mail" to reply by email. Checked
infrequently.

Your LOL is well taken. I'm not sure whether one needs the numbers
in
between the fractions (like sqrt(2)) for woodworking, nor any
negative
numbers, imaginary numbers, non-real complex numbers, nor probably
any
numbers bigger than 500. Maybe that's why those aren't marked on
the ruler.


Bill


That's why I suggested getting a sashigane marked with for
shaku/sun/bu to begin with. Once you get used to it, it has the
markings on the back side for laying out this kind of stuff without a
lot of fuss.

Len