For a given length of fine copper wire of diameter 0.072 mm (2.9 mil) =
0.004 sq mm,
if it is squashed to cross-section dimensions of 0.02 * 0.2 mm (2 * 20 mil)
proportionally how much does the resistance change ?
and then to 0.01 * 0.4mm (1 * 40 mil) ?



--
Diverse Devices, Southampton, England
electronic hints and repair briefs , schematics/manuals list on
http://home.graffiti.net/diverse:graffiti.net/

"N_Cook" wrote in message
...
For a given length of fine copper wire of diameter 0.072 mm (2.9 mil) =
0.004 sq mm,
if it is squashed to cross-section dimensions of 0.02 * 0.2 mm (2 * 20
mil)
proportionally how much does the resistance change ?
and then to 0.01 * 0.4mm (1 * 40 mil) ?





AFAIK the resistance of wire is proportional to its Cross Sectional Area.
Period. If this remains unchanged, so does the resistance.



Gareth.

"Gareth Magennis" wrote in message
...

"N_Cook" wrote in message
...
For a given length of fine copper wire of diameter 0.072 mm (2.9 mil) =
0.004 sq mm,
if it is squashed to cross-section dimensions of 0.02 * 0.2 mm (2 * 20 mil)
proportionally how much does the resistance change ?
and then to 0.01 * 0.4mm (1 * 40 mil) ?





AFAIK the resistance of wire is proportional to its Cross Sectional Area.
Period. If this remains unchanged, so does the resistance.



Gareth.



That is correct, but the length also has to remain unchanged The formula for
the resistance of a conductor is
R=r*L/A
where R= Resistance
r=Resistivity of the conductor (1.7x10^-8 for copper)
L=Length
A=cross section area

As you can see, the resistance remains constant as long as L and A remain the
same, or change in a manner that produces the same ratio.

--
Dave M
MasonDG44 at comcast dot net (Just substitute the appropriate characters in the
address)

Life is like a roll of toilet paper; the closer it gets to the end, the faster
it goes.

"DaveM" wrote in message
...
"Gareth Magennis" wrote in message
...

"N_Cook" wrote in message
...
For a given length of fine copper wire of diameter 0.072 mm (2.9 mil) =
0.004 sq mm,
if it is squashed to cross-section dimensions of 0.02 * 0.2 mm (2 * 20
mil)
proportionally how much does the resistance change ?
and then to 0.01 * 0.4mm (1 * 40 mil) ?





AFAIK the resistance of wire is proportional to its Cross Sectional Area.
Period. If this remains unchanged, so does the resistance.



Gareth.



That is correct, but the length also has to remain unchanged The formula
for the resistance of a conductor is
R=r*L/A
where R= Resistance
r=Resistivity of the conductor (1.7x10^-8 for copper)
L=Length
A=cross section area

As you can see, the resistance remains constant as long as L and A remain
the same, or change in a manner that produces the same ratio.

--


So that begs the question, how much can a piece of copper wire be
compressed? If you do squash it into a different shape, does or can its
volume change significantly?


Gareth.

Gareth Magennis wrote in message
...

"DaveM" wrote in message
...
"Gareth Magennis" wrote in message
...

"N_Cook" wrote in message
...
For a given length of fine copper wire of diameter 0.072 mm (2.9 mil)
=
0.004 sq mm,
if it is squashed to cross-section dimensions of 0.02 * 0.2 mm (2 * 20
mil)
proportionally how much does the resistance change ?
and then to 0.01 * 0.4mm (1 * 40 mil) ?





AFAIK the resistance of wire is proportional to its Cross Sectional
Area.
Period. If this remains unchanged, so does the resistance.



Gareth.



That is correct, but the length also has to remain unchanged The
formula
for the resistance of a conductor is
R=r*L/A
where R= Resistance
r=Resistivity of the conductor (1.7x10^-8 for copper)
L=Length
A=cross section area

As you can see, the resistance remains constant as long as L and A
remain
the same, or change in a manner that produces the same ratio.

--


So that begs the question, how much can a piece of copper wire be
compressed? If you do squash it into a different shape, does or can its
volume change significantly?


Gareth.



So it may be an effect of work hardening , relative increase in the effect
of imperfections/micro fractures or some other metallurgical effect.
Mackie speaker voice coil failures due to this flattening/ribboning process
to make the tails to the outside world.
Previous failure at the juncture of round to flat (0.07mm round to about
0.02 x 0.2mm) so at the peak stress point.
This one along the length of the ribbon section, but the whole 50mm or so
run was brittleised and disintegrated on touch, not the slightest sign of
overheating on the remaining 25 turns of round wire.
broken end marked B on this pic
http://home.graffiti.net/diverse:gra...ckie_horn1.jpg
http://home.graffiti.net/diverse:gra...ckie_horn2.jpg
Cannot expore the metallurgy as that curve of "wire" as totally
disintegrated to dust.


--
Diverse Devices, Southampton, England
electronic hints and repair briefs , schematics/manuals list on
http://home.graffiti.net/diverse:graffiti.net/

"Gareth Magennis" wrote in message
...

"DaveM" wrote in message
...
"Gareth Magennis" wrote in message
...

"N_Cook" wrote in message
...
For a given length of fine copper wire of diameter 0.072 mm (2.9 mil) =
0.004 sq mm,
if it is squashed to cross-section dimensions of 0.02 * 0.2 mm (2 * 20 mil)
proportionally how much does the resistance change ?
and then to 0.01 * 0.4mm (1 * 40 mil) ?





AFAIK the resistance of wire is proportional to its Cross Sectional Area.
Period. If this remains unchanged, so does the resistance.



Gareth.



That is correct, but the length also has to remain unchanged The formula for
the resistance of a conductor is
R=r*L/A
where R= Resistance
r=Resistivity of the conductor (1.7x10^-8 for copper)
L=Length
A=cross section area

As you can see, the resistance remains constant as long as L and A remain the
same, or change in a manner that produces the same ratio.

--


So that begs the question, how much can a piece of copper wire be compressed?
If you do squash it into a different shape, does or can its volume change
significantly?


Gareth.


The shape of the cross section can change to virtually any dimension so long as
the length remains the same. IOW, if you squeeze a bar of 10mmx10mm down to
2mmx50mm, its cross sectional area stayed constant (only the shape of the area
changed). Its length will remain the same, since the volume didn't change;
hence, its resistance will remain the same.
So long as material is not added or removed, the volume will remain the same.
The formula says that the ratio of length to cross-sectional area must remain
the same in order for resistance to remain unchanged. If cross sectional area
is changed, the length must change to maintain the ratio. The volume must
remain constant.

--
Dave M
MasonDG44 at comcast dot net (Just substitute the appropriate characters in the
address)

Life is like a roll of toilet paper; the closer it gets to the end, the faster
it goes.

On Dec 12, 11:55*am, "DaveM" wrote:
"Gareth Magennis" wrote in message

...







"DaveM" wrote in message
...
"Gareth Magennis" wrote in message
...

"N_Cook" wrote in message
...
For a given length of fine copper wire of diameter 0.072 mm (2.9 mil) =
0.004 sq mm,
if it is squashed to cross-section dimensions of 0.02 * 0.2 mm (2 * 20 mil)
proportionally how much does the resistance change ?
and then to 0.01 * 0.4mm (1 * 40 mil) ?

AFAIK the resistance of wire is proportional to its Cross Sectional Area.
Period. *If this remains unchanged, so does the resistance.

Gareth.

That is correct, but the length also has to remain unchanged *The formula for
the resistance of a conductor is
R=r*L/A
where R= Resistance
r=Resistivity of the conductor (1.7x10^-8 for copper)
L=Length
A=cross section area

As you can see, the resistance remains constant as long as L and A remain the
same, or change in a manner that produces the same ratio.

--

So that begs the question, how much can a piece of copper wire be compressed?
If you do squash it into a different shape, does or can its volume change
significantly?

Gareth.

The shape of the cross section can change to virtually any dimension so long as
the length remains the same. *IOW, if you squeeze a bar of 10mmx10mm down to
2mmx50mm, its cross sectional area stayed constant (only the shape of the area
changed). *Its length will remain the same, since the volume didn't change;
hence, its resistance will remain the same.
So long as material is not added or removed, the volume will remain the same.
The formula says that the ratio of length to cross-sectional area must remain
the same in order for resistance to remain unchanged. *If cross sectional area
is changed, the length must change to maintain the ratio. *The volume must
remain constant.

--
Dave M
MasonDG44 at comcast dot net *(Just substitute the appropriate characters in the
address)

Life is like a roll of toilet paper; the closer it gets to the end, the faster
it goes.- Hide quoted text -

- Show quoted text -

What I believe Norm is questioning/proposing is that the wire may have
been made more "dense" by being compressed without lengthening, and
that would probably decrease its resistance.

Bob Hofmann