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Fergus Kyle
Joined: 03 Jun 2007 Posts: 291 Location: Burlington ON Canada
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Posted: Sat Sep 06, 2008 12:16 pm Post subject: What did I do wrong? |
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Well, of course - my memory of simple math let me down again!
Thanks to rjquillin and N8ZG, I have found that pi*D is the circumference.
On the other hand I got three different answers, since Circular Mils is a
square of D, the answers are a bit different.
In any event, what an advantage this net is!
Thanks again!
Ferg
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rjquillin
Joined: 13 May 2007 Posts: 123 Location: KSEE
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Posted: Sat Sep 06, 2008 3:16 pm Post subject: What did I do wrong? |
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At 13:14 9/6/2008, you wrote:
Quote: | Well, of course - my memory of simple math let me down again!
Thanks to rjquillin and N8ZG, I have found that pi*D is the circumference.
On the other hand I got three different answers, since Circular Mils is a
square of D, the answers are a bit different.
In any event, what an advantage this net is!
Thanks again!
Ferg
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Mercy, the coffee and day finally kicked in...
No one ripped me earlier, despite my rightfully deserving it.
Besides being decimally challenged, and suffering fat fingers, I
forgot the CMA to in2 conversion of .7854 x 10e-6.
That .204 diameter listed is for solid wire, you don't have solid
wire and there is void area between the individual strands of the
cable, so to use that, solid, diameter would yield a result with some error.
You said #4 welding so it's likely not the 7x19/25 rope lay cable,
but either 259/27 or more likely 1666/36. The CMA of 1666/36 is
41650 from a stranded wire table. Cross sectional area in sq mils =
CMA x .7854; 41650 x .7854 = 32712 square mil or 0.032712 square
in. So .032712/.125 = 0.261696 width required to equal the same
conductor area.
Factoring in resistivity, from an old Chem-Physics handbook,
'properties of metals used as conductors' lists resistivity of
annealed copper as 1.72 and brass as 7. So brass has about 4.12
times more resistance. Take that .26 x 4.12 and you have a bit over
an inch; 1.07.
So, if I haven't again had a massive brain f**t, it would seem a bit
more than an inch width of your 1/8 inch brass would equal AWG-4
copper welding cable; at 20C.
I think I got it correct this time, but do check it out...
Ron Q.
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nuckolls.bob(at)cox.net Guest
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Posted: Sun Sep 07, 2008 11:57 am Post subject: What did I do wrong? |
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At 04:13 PM 9/6/2008 -0700, you wrote:
Quote: |
At 13:14 9/6/2008, you wrote:
>Well, of course - my memory of simple math let me down again!
>Thanks to rjquillin and N8ZG, I have found that pi*D is the circumference.
>On the other hand I got three different answers, since Circular Mils is a
>square of D, the answers are a bit different.
> In any event, what an advantage this net is!
>Thanks again!
>Ferg
Mercy, the coffee and day finally kicked in...
No one ripped me earlier, despite my rightfully deserving it.
Besides being decimally challenged, and suffering fat fingers, I forgot
the CMA to in2 conversion of .7854 x 10e-6.
That .204 diameter listed is for solid wire, you don't have solid wire and
there is void area between the individual strands of the cable, so to use
that, solid, diameter would yield a result with some error.
You said #4 welding so it's likely not the 7x19/25 rope lay cable, but
either 259/27 or more likely 1666/36. The CMA of 1666/36 is 41650 from a
stranded wire table. Cross sectional area in sq mils = CMA x .7854;
41650 x .7854 = 32712 square mil or 0.032712 square in. So .032712/.125 =
0.261696 width required to equal the same conductor area.
Factoring in resistivity, from an old Chem-Physics handbook, 'properties
of metals used as conductors' lists resistivity of annealed copper as 1.72
and brass as 7. So brass has about 4.12 times more resistance. Take that
.26 x 4.12 and you have a bit over an inch; 1.07.
So, if I haven't again had a massive brain f**t, it would seem a bit more
than an inch width of your 1/8 inch brass would equal AWG-4 copper welding
cable; at 20C.
I think I got it correct this time, but do check it out...
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Bravo my friend. There is no substitute for considered
application of simple ideas which includes facts of physics
and artfully applied math. Your assumptions and deductions
based thereon are in the ballpark. I'll leave it to others
to grade your math. The point I'd like to make here is
that all such questions yield nicely to this kind
of thought process. It's also useful to consider other
details. For example, I've oft suggested that ALL battery(+)
and battery(-) jumpers be made of 4AWG welding cable irrespective
of fat wire sizes used elsewhere. This is a consideration of
the fragile nature of lead posts used on some batteries
combined with the difficulty of using only two hands to
position wire, bolt, and washer in relation to the post
while a nut is installed. More than one set of hardware
has been launched into the nether regions of structure
by a stiff battery jumper that got loose!
It's a study in trade offs. If the electrics of Ferg's jumper
are higher loss than 4AWG equivalent, how bad is the temperature
rise during a cranking event? Keep in mind that the flat strap
sheds heat faster than a round, insulated wire. How bad is
the voltage drop during a cranking event? We KNOW that the
system total loop resistance is significant. See:
http://aeroelectric.com/Pictures/Schematics/Voltage_Drop_Study_2.pdf
Gee . . . it may well be that the thin brass strip originally
proposed might be okay in the grand scheme of things.
I mention all this only because the elegant solution
considers ALL the ingredients that go into a recipe for
success . . . of which getting the math right for deducing
equivalency is but one part.
Thanks for taking the time to walk us through this
exercise!
Bob . . .
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Fergus Kyle
Joined: 03 Jun 2007 Posts: 291 Location: Burlington ON Canada
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Posted: Mon Sep 08, 2008 6:33 am Post subject: What did I do wrong? |
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Bob,
You said:
" It's a study in trade offs. If the electrics of Ferg's jumper
are higher loss than 4AWG equivalent, how bad is the temperature
rise during a cranking event? Keep in mind that the flat strap
sheds heat faster than a round, insulated wire. How bad is
the voltage drop during a cranking event? We KNOW that the
system total loop resistance is significant. See:
http://aeroelectric.com/Pictures/Schematics/Voltage_Drop_Study_2.pdf
Gee . . . it may well be that the thin brass strip originally
proposed might be okay in the grand scheme of things.
I mention all this only because the elegant solution
considers ALL the ingredients that go into a recipe for
success . . . of which getting the math right for deducing
equivalency is but one part."
Well, I went there, and behold a neat resistance diagram - good
stuff...... Did you do this with a "megger"? I remember my mentor (lo
those many years) going over all my work with his megger to confirm - I
thought - good connections and no shorts. Is that still the routine?
Ferg
PS: My main purpose in brass was to establish the shape and size while
looking for a copper source.
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rjquillin
Joined: 13 May 2007 Posts: 123 Location: KSEE
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Posted: Mon Sep 08, 2008 7:04 am Post subject: What did I do wrong? |
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At 07:26 9/8/2008, you wrote:
Quote: | PS: My main purpose in brass was to establish the shape and size while
looking for a copper source.
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Try McMaster-Carr.
8964K271 is a 0.125 x 12" Alloy 110 Electronic-Grade Copper for $7.33.
Many other sizes available.
http://www.mcmaster.com/
Ron Q.
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Fergus Kyle
Joined: 03 Jun 2007 Posts: 291 Location: Burlington ON Canada
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Posted: Mon Sep 08, 2008 8:37 am Post subject: What did I do wrong? |
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Cheers,
At the risk of flogging this dog of a topic, I consulted an old
friend, "Handbook of
Applied Mathematics" in stumbling through several attempts to assimilate
Ron's excellent
application. I quote:
"Wires are often grouped in smaller ropes for ease of use. The group of
wires is called a
"strand"; the term "wire" being reserved for the individual wires of the
strand. Strands are
usually built of wires of such a size that the cross-section of the metal in
the strand is the
same as the cross-section of the solid wire having the same gage (sic)
number."
This book also gave me a value of 204 mils in diam.
Half that, squared, times pi, gave me 32781 square mils. Divide by a
thickness of
125 mils gives 261-odd mils or ¼ inch. So, .25 by .125 inches seems correct
but 1/16 by
½ inch seems easier, in copper.
I must remember pie are round, cake are square.
Ferg
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nuckolls.bob(at)cox.net Guest
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Posted: Mon Sep 08, 2008 6:32 pm Post subject: What did I do wrong? |
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At 10:26 AM 9/8/2008 -0400, you wrote:
Quote: | Bob,
You said:
" It's a study in trade offs. If the electrics of Ferg's jumper
are higher loss than 4AWG equivalent, how bad is the temperature
rise during a cranking event? Keep in mind that the flat strap
sheds heat faster than a round, insulated wire. How bad is
the voltage drop during a cranking event? We KNOW that the
system total loop resistance is significant. See:
http://aeroelectric.com/Pictures/Schematics/Voltage_Drop_Study_2.pdf
Gee . . . it may well be that the thin brass strip originally
proposed might be okay in the grand scheme of things.
I mention all this only because the elegant solution
considers ALL the ingredients that go into a recipe for
success . . . of which getting the math right for deducing
equivalency is but one part."
|
Quote: | Well, I went there, and behold a neat resistance diagram - good
stuff...... Did you do this with a "megger"? I remember my mentor (lo
those many years) going over all my work with his megger to confirm - I
thought - good connections and no shorts. Is that still the routine?
Ferg
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A 'megger' is what the name implies, a meg-ohmmeter
usually with a capability of reading thousands of
meg ohms. This device would be used to measure insulation
resistance while powered up at hundreds if not 1000 volts.
The diagram was a hypothetical based on experiences. Although
there are micro-ohmmeters that can be used to measure very
small resistances directly. The one I have is like this
http://www.avtron.com/pdf/ate/t477w.pdf
While the 'megger' uses very high voltage to detect and
quantify very large resistances the micro-ohmmeter
or bonding meter uses high currents to detect and
quantify very low resistances.
Quote: | PS: My main purpose in brass was to establish the shape and size while
looking for a copper source.
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Did we get you covered? There was a lot of 'stuff' flying
around there for awhile. 4awg is 0.2 dia, 0.1 radius.
0.1(squared) x 3.14 or 0.031 square inches cross section. This
means that a .032" piece 1" wide copper has the same cross
section and conductivity.
I may have some .032" or 22 gage (.025") sheet stock
in M.L. I'll check tomorrow when I'm there.
Bob . . .
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