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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 11:06 am Post subject: What am I doing wrong? |
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I know, I know.
I get into these pickles and need a rescue - must be age.....
I want to replace a #4SWG welding cable with brass bar (copper not
available).
The diameter of #4 is (AE Connection, chap. 204 mils, or 0.204".
That, times
pi, should give the area. Divide the area by 1/8inch thick bar, gives 5127.1
mils, or
5 inches. I cannot for the life of me believe that a bar cross-section of 5
by 1/8 inches
Is equivalent (approximately) to this piece of #4 here in my hands.....
What am I doing wrong?
Ferg
$1/10 =10 cents; square both sides; $1/100 = 100 cents; 1 cent = $1.
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rjquillin
Joined: 13 May 2007
Posts: 123
Location: KSEE
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| Posted: Sat Sep 06, 2008 11:31 am Post subject: What am I doing wrong? |
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At 12:04 9/6/2008, you wrote:
| Quote: | I know, I know.
I get into these pickles and need a rescue - must be age.....
I want to replace a #4SWG welding cable with brass bar (copper not
available).
The diameter of #4 is (AE Connection, chap. 204 mils, or 0.204".
That, times
pi, should give the area.
|
Actially area is pi x radius squared. But why not just look up the
cross-sectional area?
Depending on stranding AWG-4 the wire table I have shows 41650 to
53314 CMA or 0.232 to 0.257 inch.
Using .257 that would give an area of .40369 sq. in.
| Quote: | Divide the area by 1/8inch thick bar, gives 5127.1
mils, or
5 inches. I cannot for the life of me believe that a bar cross-section of 5
by 1/8 inches
Is equivalent (approximately) to this piece of #4 here in my hands.....
What am I doing wrong?
|
If you call the area 0.4 and divide by .125 you get 3.2 inches.
Still wide...
| Quote: | Ferg
$1/10 =10 cents; square both sides; $1/100 = 100 cents; 1 cent = $1.
|
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http://www.matronics.com/Navigator?AeroElectric-List |
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rjquillin
Joined: 13 May 2007
Posts: 123
Location: KSEE
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| Posted: Sat Sep 06, 2008 11:33 am Post subject: What am I doing wrong? |
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Oops, some edits got left out...
At 12:04 9/6/2008, you wrote:
| Quote: | I know, I know.
I get into these pickles and need a rescue - must be age.....
I want to replace a #4SWG welding cable with brass bar (copper not
available).
The diameter of #4 is (AE Connection, chap. 204 mils, or
0.204" diameter.
That, times
pi, should give the area.
|
Pi x diameter is circumference.
Actially area is pi x radius squared. But why not just look up the
cross-sectional area?
Depending on stranding AWG-4 the wire table I have shows 41650 to
53314 CMA or 0.232 to 0.257 inch.
Using .257 that would give an area of .40369 sq. in.
| Quote: | Divide the area by 1/8inch thick bar, gives 5127.1
mils, or
5 inches. I cannot for the life of me believe that a bar cross-section of 5
by 1/8 inches
Is equivalent (approximately) to this piece of #4 here in my hands.....
What am I doing wrong?
|
If you call the area 0.4 and divide by .125 you get 3.2 inches wide.
Using .53 area you get 4.24 inches wide.
Still pretty wide...
| Quote: | Ferg
$1/10 =10 cents; square both sides; $1/100 = 100 cents; 1 cent = $1.
|
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n8zg(at)mchsi.com
Guest
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| Posted: Sat Sep 06, 2008 11:43 am Post subject: What am I doing wrong? |
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No…
Circumference is pi*d
Area is pi*r(squared)
So… if d=0.204
Then r=0.102
And a=pi*0.102*0.102
a=3.14159*0.010404
a=0.03269 square inches
a/t=w
0.03269/0.125 = 0.2179 wide
Might as well say 1/8” x 1/4”
But you need enough to put a bolt thru.
Cut it 5/8” wide.
neal
_____________________________________________
From: owner-aeroelectric-list-server(at)matronics.com [mailto:owner-aeroelectric-list-server(at)matronics.com (owner-aeroelectric-list-server(at)matronics.com)] On Behalf Of Fergus Kyle
Sent: Saturday, September 06, 2008 2:05 PM
To: aeroelectric-list(at)matronics.com
Subject: What am I doing wrong?
I know, I know.
I get into these pickles and need a rescue – must be age…..
I want to replace a #4SWG welding cable with brass bar (copper not available).
The diameter of #4 is (AE Connection, chap. 204 mils, or 0.204”. That, times
pi, should give the area. Divide the area by 1/8inch thick bar, gives 5127.1 mils, or
5 inches. I cannot for the life of me believe that a bar cross-section of 5 by 1/8 inches
Is equivalent (approximately) to this piece of #4 here in my hands…..
What am I doing wrong?
Ferg
$1/10 =10 cents; square both sides; $1/100 = 100 cents; 1 cent = $1.
[quote][b]
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rv8ch
Joined: 10 Jan 2006
Posts: 250
Location: Switzerland
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| Posted: Sat Sep 06, 2008 11:49 am Post subject: What am I doing wrong? |
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Ferg,
Your math is wrong. Area of a circle is pi * r^2
Check this link for details:
http://en.wikipedia.org/wiki/American_wire_gauge
Mickey
Fergus Kyle wrote:
| Quote: | I know, I know.
I get into these pickles and need a rescue - must be age.....
I want to replace a #4SWG welding cable with brass bar (copper not
available).
The diameter of #4 is (AE Connection, chap. 204 mils, or 0.204".
That, times
pi, should give the area. Divide the area by 1/8inch thick bar, gives 5127.1
mils, or
5 inches. I cannot for the life of me believe that a bar cross-section of 5
by 1/8 inches
Is equivalent (approximately) to this piece of #4 here in my hands.....
What am I doing wrong?
Ferg
$1/10 =10 cents; square both sides; $1/100 = 100 cents; 1 cent = $1.
|
--
Mickey Coggins
http://www.rv8.ch/
#82007 finishing
do not archive
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_________________
Mickey Coggins
http://www.rv8.ch/ |
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JohnInReno
Joined: 08 Sep 2007
Posts: 150
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| Posted: Sat Sep 06, 2008 11:59 am Post subject: What am I doing wrong? |
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Just a guess, but isn't the formula for area = pi x radius squared?
.204/2 = .102
.102 x .102 = .0104
.0104 x 3.1416 = .0327
.0327 / .125 = .261 or just over 1/4"
john
Fergus Kyle wrote:
| Quote: | I know, I know.
I get into these pickles and need a rescue - must be age.....
I want to replace a #4SWG welding cable with brass bar (copper not
available).
The diameter of #4 is (AE Connection, chap. 204 mils, or 0.204".
That, times
pi, should give the area. Divide the area by 1/8inch thick bar, gives 5127.1
mils, or
5 inches. I cannot for the life of me believe that a bar cross-section of 5
by 1/8 inches
Is equivalent (approximately) to this piece of #4 here in my hands.....
What am I doing wrong?
Ferg
$1/10 =10 cents; square both sides; $1/100 = 100 cents; 1 cent = $1.
|
| | - The Matronics AeroElectric-List Email Forum - | | | Use the List Feature Navigator to browse the many List utilities available such as the Email Subscriptions page, Archive Search & Download, 7-Day Browse, Chat, FAQ, Photoshare, and much more:
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John Morgensen
RV-9A - Born on July 3, 2013
RV4 - for sale |
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mjpienaar(at)shaw.ca
Guest
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| Posted: Sat Sep 06, 2008 11:59 am Post subject: What am I doing wrong? |
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Formula is pi * r*r
3.141 * .102 * .102 = .032685
.032685/.125 = .2614
Thus width of bar required is +/- 1/4 inch.
I'm not an angineer but hope I'm right, good luck
Mike
---
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rv-9a-online(at)telus.net
Guest
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| Posted: Sat Sep 06, 2008 12:32 pm Post subject: What am I doing wrong? |
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Not to be a pita, but brass has several times the resistance of copper, depending on the alloy.
Van's Aircraft sells it (ES BUSS BAR-063X.5X12) which is just about what you need.
If you use brass, you'd need a lot wider or thicker strip (2 to 7 times depending on alloy).
Vern Little
--
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lm4(at)juno.com
Guest
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| Posted: Sun Sep 07, 2008 7:29 am Post subject: What am I doing wrong? |
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Fergus,
Area is Pi-R squared. Not Pi-Dia.
On Sat, 6 Sep 2008 15:04:37 -0400 "Fergus Kyle" <VE3LVO(at)rac.ca> writes:
| Quote: | I know, I know.
I get into these pickles and need a rescue - must be
age.....
I want to replace a #4SWG welding cable with brass bar
(copper not
available).
The diameter of #4 is (AE Connection, chap. 204 mils, or
0.204".
That, times
pi, should give the area. Divide the area by 1/8inch thick bar,
gives 5127.1
mils, or
5 inches. I cannot for the life of me believe that a bar
cross-section of 5
by 1/8 inches
Is equivalent (approximately) to this piece of #4 here in my
hands.....
What am I doing wrong?
Ferg
$1/10 =10 cents; square both sides; $1/100 = 100 cents; 1 cent =
$1.
|
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