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What did I do wrong?

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Fergus Kyle

Joined: 03 Jun 2007
Posts: 291
Location: Burlington ON Canada

Posted: Sat Sep 06, 2008 12:16 pm Post subject: What did I do wrong?

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

Posted: Sat Sep 06, 2008 3:16 pm Post subject: What did I do wrong?

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


	


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

Posted: Sun Sep 07, 2008 11:57 am Post subject: What did I do wrong?

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...


	


    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

Posted: Mon Sep 08, 2008 6:33 am Post subject: What did I do wrong?

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

Posted: Mon Sep 08, 2008 7:04 am Post subject: What did I do wrong?

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.


	


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

Posted: Mon Sep 08, 2008 8:37 am Post subject: What did I do wrong?

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
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Posted: Mon Sep 08, 2008 6:32 pm Post subject: What did I do wrong?

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


	


    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.


	


    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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