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Date: Tue, 24 Nov 2009 14:29:21 -0800 (PST)
From: Dwayne Parkinson <dwayneparkinson@yahoo.com>
Subject: Re: [FlyRotary] Re: Typical Injector Current Draw
To: Rotary motors in aircraft <flyrotary@lancaironline.net>
In-Reply-To: <list-3988386@logan.com>
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Wow.  Ed, you are one amazing dude.=0A=0A=0A=0A____________________________=
____=0AFrom: Ed Anderson <eanderson@carolina.rr.com>=0ATo: Rotary motors in=
 aircraft <flyrotary@lancaironline.net>=0ASent: Tue, November 24, 2009 2:44=
:52 PM=0ASubject: [FlyRotary] Re: Typical Injector Current Draw=0A=0A =0AHi=
 Jeff,=0A =0ASince your Renesis injector is the saturated type, it=0Aprobab=
ly has a resistance on the order of 12 ohms (could be anywhere between=0Aar=
ound 8 =E2=80=93 16 ohms =E2=80=93 would have to measure it to be certain.=
=0A =0AGiven a 13.8 Volt system that would mean it would draw I =3D=0AE/R =
=3D 13.8/12 =3D 1.5 amps when there is sufficient current to pull it=0Aopen=
.  Perhaps a bit more or a bit less.  The dynamics of current=0Achanging in=
 a magnetic field makes this a non-linear relationship =E2=80=93 but=0Afor =
a back of the envelope approach lets ignore that little detail.  Its=0Aunli=
kely the current draw would be more.  =0A =0AThis shows the current vs time=
 for a typical saturation type=0Ainjector.  As you can see the current appe=
ars pretty much proportional to=0Athe amount of time the injector is open u=
p to the point it stabilizes.  Which=0AI think would be around 1 msec (but =
just a guess in this case)=0A =0AAt WOT at 6000 rpm the maximum PW would be=
 10 msec (because=0Athat happens to be the injection cycle for the rotary a=
t 6000 rpm).  So at=0Athat point the injector would be open/on continuously=
.  So lets consider=0Athat the maximum power draw case =E2=80=93 whenever t=
he injectors are=0A=E2=80=9Clocked Open=E2=80=9D =E2=80=93 you don=E2=80=99=
t=E2=80=99 want to operate there=0A{:>).  But, your one injector would be o=
pen continuously - drawing 1.5=0Aamps.  =0A =0AHowever, that would equate t=
o the engine consuming 29.16 GPH=0A(4 x 460 cc/min injectors) which is high=
ly unlikely at 6000 rpm and no boost=0A=E2=80=93 your engine would probably=
 bog down and flood at that flow rate at 6000=0Arpm.=0ASo something less th=
an that is probably the case..=0A =0ALet=E2=80=99s say our engine will =E2=
=80=9Crealistically=E2=80=9D=0Aproduce 200 HP at 7500 rpm that would equate=
 to approx 20.0 gph fuel flow using=0Aa BSFC of 0.55 to make the calculatio=
n.=0A =0AFor 4 x 460 cc/min injectors to give you that fuel=0Aflow  at 7500=
 rpm would require them to be open approx 5.1 msec=0AThe injection period a=
t 7500 rpm is around 8 msec (so we=0Ahave not =E2=80=9Clock open=E2=80=9D t=
he injectors at this point)=0A =0ASo the duty cycle would be 5.1/8 =3D  64%=
=0A =0ATherefore if one injector draws 1.5 amps when open all the=0Atime th=
en 4 would draw 4 x 1.5 =3D  6 amps=0A =0ASo with a duty cycle of=0A64% the=
y would have  a combined current draw of approx 6 * 0.64 =3D 3.825=0Aamps a=
t 7500 rpm.  Now as the power requirement decreases the injectors do=0Anot =
have to stay on as long and so your long-term current draw is undoubtedly=
=0Aless than this =E2=80=93 unless you run WOT all the time {:>)=0ANow if y=
ou used smaller injectors the injectors=0Awould have to be open longer drai=
ng more current (for that 200 hp) and if=0Alarger injectors then they would=
 be open less time drawing less power due to=0Atheir lower duty cycle.=0AWe=
ll, at least that is the way it seems to me=0Afrom the back of the envelope=
.=0A =0AEd=0A =0AEd Anderson=0ARv-6A N494BW Rotary Powered=0AMatthews, NC=
=0Aeanderson@carolina.rr.com=0Ahttp://www.andersonee.com=0Ahttp://www.dmack=
.net/mazda/index.html=0Ahttp://www.flyrotary.com/=0Ahttp://members.cox.net/=
rogersda/rotary/configs.htm#N494BW=0Ahttp://www.rotaryaviation.com/Rotorhea=
d%20Truth.htm=0A=0A________________________________=0A =0AFrom:Rotary motor=
s in aircraft [mailto:flyrotary@lancaironline.net] On Behalf Of Jeff Luckey=
=0ASent: Tuesday, November 24, 2009=0A1:09 PM=0ATo: Rotary motors in aircra=
ft=0ASubject: [FlyRotary] Typical=0AInjector Current Draw=0A =0AI=E2=80=99m=
=0Adoing some electrical load analysis and need some stats for fuel injecto=
rs.=0A =0AWhat=0Adoes a typical fuel injector used on a Renesis draw?  I=E2=
=80=99m looking=0Afor 3 numbers:=0A=091. If I=E2=80=99ve got an injector on=
 the bench, what is the energized load?=0A=092. When the injector is runnin=
g in the engine, it is getting a PWM signal.  What it the typical duty cycl=
e at full throttle?  This will help calculate average load over time.  =0A=
=093. Or - is there some rule-of-thumb number that is used for engine at fu=
ll throttle?=0A =0ATIA=0A =0A-Jeff=0A =0A =0AJeff=0ALuckey=0ADirect=0AConne=
ction Systems=0Awww.DCSIT.com=0A949-645-8832=0A =0A=0A=0A__________ Informa=
tion from ESET NOD32 Antivirus, version of virus signature=0Adatabase 3267 =
(20080714) __________=0A=0AThe message was checked by ESET NOD32 Antivirus.=
=0A=0Ahttp://www.eset.com=0A=0A=0A__________ Information from ESET NOD32 An=
tivirus, version of virus signature=0Adatabase 3267 (20080714) __________=
=0A=0AThe message was checked by ESET NOD32 Antivirus.=0A=0Ahttp://www.eset=
.com=0A=0A__________ Information from ESET NOD32 Antivirus, version of viru=
s signature database 3267 (20080714) __________=0A=0AThe message was checke=
d by ESET NOD32 Antivirus.=0A=0Ahttp://www.eset.com=0A=0A=0A=0A      
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<html><head><style type=3D"text/css"><!-- DIV {margin:0px;} --></style></he=
ad><body><div style=3D"font-family:arial, helvetica, sans-serif;font-size:1=
0pt"><div>Wow. &nbsp;Ed, you are one amazing dude.</div><div style=3D"font-=
family:arial, helvetica, sans-serif;font-size:10pt"><br><div style=3D"font-=
family:times new roman, new york, times, serif;font-size:12pt"><font size=
=3D"2" face=3D"Tahoma"><hr size=3D"1"><b><span style=3D"font-weight: bold;"=
>From:</span></b> Ed Anderson &lt;eanderson@carolina.rr.com&gt;<br><b><span=
 style=3D"font-weight: bold;">To:</span></b> Rotary motors in aircraft &lt;=
flyrotary@lancaironline.net&gt;<br><b><span style=3D"font-weight: bold;">Se=
nt:</span></b> Tue, November 24, 2009 2:44:52 PM<br><b><span style=3D"font-=
weight: bold;">Subject:</span></b> [FlyRotary] Re: Typical Injector Current=
 Draw<br></font><br>=0A=0A=0A=0A =0A=0A=0A =0A=0A =0A=0A<style>=0A<!--=0A =
=0A _filtered {font-family:Tahoma;panose-1:2 11 6 4 3 5 4 4 2 4;}=0A =0Ap.M=
soNormal, li.MsoNormal, div.MsoNormal=0A=09{margin:0in;margin-bottom:.0001p=
t;font-size:9.0pt;font-family:Arial;}=0Aa:link, span.MsoHyperlink=0A=09{col=
or:blue;text-decoration:underline;}=0Aa:visited, span.MsoHyperlinkFollowed=
=0A=09{color:purple;text-decoration:underline;}=0Ap=0A=09{margin-right:0in;=
margin-left:0in;font-size:12.0pt;font-family:"Times New Roman";}=0Aspan.Ema=
ilStyle18=0A=09{font-family:Arial;color:windowtext;}=0Aspan.EmailStyle19=0A=
=09{font-family:Arial;color:navy;}=0A _filtered {margin:1.0in 1.25in 1.0in =
1.25in;}=0Adiv.Section1=0A=09{}=0A =0A _filtered {}=0A _filtered {}=0A _fil=
tered {}=0A _filtered {}=0A _filtered {}=0A _filtered {}=0A _filtered {}=0A=
 _filtered {}=0A _filtered {}=0A _filtered {}=0A _filtered {}=0Aol=0A=09{ma=
rgin-bottom:0in;}=0Aul=0A=09{margin-bottom:0in;}=0A-->=0A</style>=0A=0A=0A=
=0A=0A=0A<div class=3D"Section1">=0A=0A<p class=3D"MsoNormal"><font size=3D=
"2" color=3D"navy" face=3D"Arial"><span style=3D"=0Afont-size:10.0pt;color:=
navy;">Hi Jeff,</span></font></p> =0A=0A<p class=3D"MsoNormal"><font size=
=3D"2" color=3D"navy" face=3D"Arial"><span style=3D"=0Afont-size:10.0pt;col=
or:navy;"> &nbsp;</span></font></p> =0A=0A<p class=3D"MsoNormal"><font size=
=3D"2" color=3D"navy" face=3D"Arial"><span style=3D"=0Afont-size:10.0pt;col=
or:navy;">Since your Renesis injector is the saturated type, it=0Aprobably =
has a resistance on the order of 12 ohms (could be anywhere between=0Aaroun=
d 8 =E2=80=93 16 ohms =E2=80=93 would have to measure it to be certain.</sp=
an></font></p> =0A=0A<p class=3D"MsoNormal"><font size=3D"2" color=3D"navy"=
 face=3D"Arial"><span style=3D"=0Afont-size:10.0pt;color:navy;"> &nbsp;</sp=
an></font></p> =0A=0A<p class=3D"MsoNormal"><font size=3D"2" color=3D"navy"=
 face=3D"Arial"><span style=3D"=0Afont-size:10.0pt;color:navy;">Given a 13.=
8 Volt system that would mean it would draw I =3D=0AE/R =3D 13.8/12 =3D 1.5=
 amps when there is sufficient current to pull it=0Aopen.&nbsp; Perhaps a b=
it more or a bit less.&nbsp; The dynamics of current=0Achanging in a magnet=
ic field makes this a non-linear relationship =E2=80=93 but=0Afor a back of=
 the envelope approach lets ignore that little detail.&nbsp; Its=0Aunlikely=
 the current draw would be more.&nbsp; <img width=3D"435" height=3D"346" id=
=3D"_x0000_i1026" src=3D"cid:1.2251235516@web46410.mail.sp1.yahoo.com"></sp=
an></font></p> =0A=0A<p class=3D"MsoNormal"><font size=3D"2" color=3D"navy"=
 face=3D"Arial"><span style=3D"=0Afont-size:10.0pt;color:navy;"> &nbsp;</sp=
an></font></p> =0A=0A<p class=3D"MsoNormal"><font size=3D"2" color=3D"navy"=
 face=3D"Arial"><span style=3D"=0Afont-size:10.0pt;color:navy;">This shows =
the current vs time for a typical saturation type=0Ainjector.&nbsp; As you =
can see the current appears pretty much proportional to=0Athe amount of tim=
e the injector is open up to the point it stabilizes.&nbsp; Which=0AI think=
 would be around 1 msec (but just a guess in this case)</span></font></p> =
=0A=0A<p class=3D"MsoNormal"><font size=3D"2" color=3D"navy" face=3D"Arial"=
><span style=3D"=0Afont-size:10.0pt;color:navy;"> &nbsp;</span></font></p> =
=0A=0A<p class=3D"MsoNormal"><font size=3D"2" color=3D"navy" face=3D"Arial"=
><span style=3D"=0Afont-size:10.0pt;color:navy;">At WOT at 6000 rpm the max=
imum PW would be 10 msec (because=0Athat happens to be the injection cycle =
for the rotary at 6000 rpm).&nbsp; So at=0Athat point the injector would be=
 open/on continuously.&nbsp; So lets consider=0Athat the maximum power draw=
 case =E2=80=93 whenever the injectors are=0A=E2=80=9Clocked Open=E2=80=9D =
=E2=80=93 you don=E2=80=99t=E2=80=99 want to operate there=0A{:&gt;).&nbsp;=
 But, your one injector would be open continuously - drawing 1.5=0Aamps.&nb=
sp; </span></font></p> =0A=0A<p class=3D"MsoNormal"><font size=3D"2" color=
=3D"navy" face=3D"Arial"><span style=3D"=0Afont-size:10.0pt;color:navy;"> &=
nbsp;</span></font></p> =0A=0A<p class=3D"MsoNormal"><font size=3D"2" color=
=3D"navy" face=3D"Arial"><span style=3D"=0Afont-size:10.0pt;color:navy;">Ho=
wever, that would equate to the engine consuming 29.16 GPH=0A(4 x 460 cc/mi=
n injectors) which is highly unlikely at 6000 rpm and no boost=0A=E2=80=93 =
your engine would probably bog down and flood at that flow rate at 6000=0Ar=
pm.</span></font></p> =0A=0A<p class=3D"MsoNormal"><font size=3D"2" color=
=3D"navy" face=3D"Arial"><span style=3D"=0Afont-size:10.0pt;color:navy;">So=
 something less than that is probably the case..</span></font></p> =0A=0A<p=
 class=3D"MsoNormal"><font size=3D"2" color=3D"navy" face=3D"Arial"><span s=
tyle=3D"=0Afont-size:10.0pt;color:navy;"> &nbsp;</span></font></p> =0A=0A<p=
 class=3D"MsoNormal"><font size=3D"2" color=3D"navy" face=3D"Arial"><span s=
tyle=3D"=0Afont-size:10.0pt;color:navy;">Let=E2=80=99s say our engine will =
=E2=80=9Crealistically=E2=80=9D=0Aproduce 200 HP at 7500 rpm that would equ=
ate to approx 20.0 gph fuel flow using=0Aa BSFC of 0.55 to make the calcula=
tion.</span></font></p> =0A=0A<p class=3D"MsoNormal"><font size=3D"2" color=
=3D"navy" face=3D"Arial"><span style=3D"=0Afont-size:10.0pt;color:navy;"> &=
nbsp;</span></font></p> =0A=0A<p class=3D"MsoNormal"><font size=3D"2" color=
=3D"navy" face=3D"Arial"><span style=3D"=0Afont-size:10.0pt;color:navy;">Fo=
r 4 x 460 cc/min injectors to give you that fuel=0Aflow&nbsp; at 7500 rpm w=
ould require them to be open approx 5.1 msec</span></font></p> =0A=0A<p cla=
ss=3D"MsoNormal"><font size=3D"2" color=3D"navy" face=3D"Arial"><span style=
=3D"=0Afont-size:10.0pt;color:navy;">The injection period at 7500 rpm is ar=
ound 8 msec (so we=0Ahave not =E2=80=9Clock open=E2=80=9D the injectors at =
this point)</span></font></p> =0A=0A<p class=3D"MsoNormal"><font size=3D"2"=
 color=3D"navy" face=3D"Arial"><span style=3D"=0Afont-size:10.0pt;color:nav=
y;"> &nbsp;</span></font></p> =0A=0A<p class=3D"MsoNormal"><font size=3D"2"=
 color=3D"navy" face=3D"Arial"><span style=3D"=0Afont-size:10.0pt;color:nav=
y;">So the duty cycle would be 5.1/8 =3D&nbsp; 64%</span></font></p> =0A=0A=
<p class=3D"MsoNormal"><font size=3D"2" color=3D"navy" face=3D"Arial"><span=
 style=3D"=0Afont-size:10.0pt;color:navy;"> &nbsp;</span></font></p> =0A=0A=
<p class=3D"MsoNormal"><font size=3D"2" color=3D"navy" face=3D"Arial"><span=
 style=3D"=0Afont-size:10.0pt;color:navy;">Therefore if one injector draws =
1.5 amps when open all the=0Atime then 4 would draw 4 x 1.5 =3D&nbsp; 6 amp=
s</span></font></p> =0A=0A<p class=3D"MsoNormal"><font size=3D"2" color=3D"=
navy" face=3D"Arial"><span style=3D"=0Afont-size:10.0pt;color:navy;"> &nbsp=
;</span></font></p> =0A=0A<p class=3D"MsoNormal"><b><font size=3D"2" color=
=3D"navy" face=3D"Arial"><span style=3D"font-size:10.0pt;color:navy;font-we=
ight:bold;">So with a duty cycle of=0A64% they would have&nbsp; a combined =
current draw of approx 6 * 0.64 =3D 3.825=0Aamps at 7500 rpm.&nbsp; Now as =
the power requirement decreases the injectors do=0Anot have to stay on as l=
ong and so your long-term current draw is undoubtedly=0Aless than this =E2=
=80=93 unless you run WOT all the time {:&gt;)</span></font></b></p> =0A=0A=
<div>=0A=0A<p><font size=3D"2" color=3D"navy" face=3D"Arial"><span lang=3D"=
EN" style=3D"font-size:10.0pt;font-family:Arial;color:navy;">Now if you use=
d smaller injectors the injectors=0Awould have to be open longer draing mor=
e current (for that 200 hp) and if=0Alarger injectors then they would be op=
en less time drawing less power due to=0Atheir lower duty cycle.</span></fo=
nt></p> =0A=0A<p><font size=3D"2" color=3D"navy" face=3D"Arial"><span lang=
=3D"EN" style=3D"font-size:10.0pt;font-family:Arial;color:navy;">Well, at l=
east that is the way it seems to me=0Afrom the back of the envelope.</span>=
</font></p> =0A=0A<p><font size=3D"2" color=3D"navy" face=3D"Arial"><span l=
ang=3D"EN" style=3D"font-size:10.0pt;font-family:Arial;color:navy;"> &nbsp;=
</span></font></p> =0A=0A<p><font size=3D"2" color=3D"navy" face=3D"Arial">=
<span lang=3D"EN" style=3D"font-size:10.0pt;font-family:Arial;color:navy;">=
Ed</span></font></p> =0A=0A<p><font size=3D"2" color=3D"navy" face=3D"Arial=
"><span lang=3D"EN" style=3D"font-size:10.0pt;font-family:Arial;color:navy;=
"> &nbsp;</span></font></p> =0A=0A<p><font size=3D"2" color=3D"navy" face=
=3D"Arial"><span lang=3D"EN" style=3D"font-size:10.0pt;font-family:Arial;co=
lor:navy;">Ed Anderson</span></font></p> =0A=0A<p><font size=3D"2" color=3D=
"navy" face=3D"Arial"><span lang=3D"EN" style=3D"font-size:10.0pt;font-fami=
ly:Arial;color:navy;">Rv-6A N494BW Rotary Powered</span></font></p> =0A=0A<=
p><font size=3D"2" color=3D"navy" face=3D"Arial"><span lang=3D"EN" style=3D=
"font-size:10.0pt;font-family:Arial;color:navy;">Matthews, NC</span></font>=
</p> =0A=0A<p><font size=3D"2" color=3D"navy" face=3D"Arial"><span lang=3D"=
EN" style=3D"font-size:10.0pt;font-family:Arial;color:navy;">eanderson@caro=
lina.rr.com</span></font></p> =0A=0A<p><font size=3D"3" face=3D"Times New R=
oman"><span style=3D"font-size:12.0pt;"><a rel=3D"nofollow" target=3D"_blan=
k" href=3D"http://www.andersonee.com"><font size=3D"2" face=3D"Arial"><span=
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sonee.com</span></font></a></span></font><font size=3D"2" color=3D"navy" fa=
ce=3D"Arial"><span style=3D"font-size:10.0pt;font-family:Arial;color:navy;"=
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ion.com/Rotorhead%20Truth.htm"><font size=3D"2" face=3D"Arial"><span style=
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rhead%20Truth.htm</span></font></a><a rel=3D"nofollow" target=3D"_blank" hr=
ef=3D"http://www.dmack.net/mazda/index.html"></a></span></font><font size=
=3D"2" face=3D"Arial"><span style=3D"font-size:10.0pt;font-family:Arial;"><=
/span></font></p> =0A=0A</div>=0A=0A<div>=0A=0A<div class=3D"MsoNormal" ali=
gn=3D"center" style=3D"text-align:center;"><font size=3D"3" face=3D"Times N=
ew Roman"><span style=3D"font-size:12.0pt;font-family:&quot;Times New Roman=
&quot;;">=0A=0A<hr size=3D"2" width=3D"100%" align=3D"center" tabindex=3D"-=
1">=0A=0A</span></font></div>=0A=0A<p class=3D"MsoNormal"><b><font size=3D"=
2" face=3D"Tahoma"><span style=3D"font-size:10.0pt;font-family:Tahoma;font-=
weight:bold;">From:</span></font></b><font size=3D"2" face=3D"Tahoma"><span=
 style=3D"font-size:10.0pt;font-family:Tahoma;">=0A Rotary motors in aircra=
ft [mailto:flyrotary@lancaironline.net]=0A<b><span style=3D"font-weight:bol=
d;">On Behalf Of </span></b>Jeff Luckey<br>=0A<b><span style=3D"font-weight=
:bold;">Sent:</span></b> Tuesday, November 24, 2009=0A1:09 PM<br>=0A<b><spa=
n style=3D"font-weight:bold;">To:</span></b> Rotary=0A motors in aircraft<b=
r>=0A<b><span style=3D"font-weight:bold;">Subject:</span></b> [FlyRotary] T=
ypical=0AInjector Current Draw</span></font><font size=3D"3" face=3D"Times =
New Roman"><span style=3D"font-size:12.0pt;font-family:&quot;Times New Roma=
n&quot;;"></span></font></p> =0A=0A</div>=0A=0A<p class=3D"MsoNormal"><font=
 size=3D"1" face=3D"Arial"><span style=3D"font-size:9.0pt;"> &nbsp;</span><=
/font></p> =0A=0A<p class=3D"MsoNormal"><font size=3D"2" face=3D"Arial"><sp=
an style=3D"font-size:10.0pt;">I=E2=80=99m=0Adoing some electrical load ana=
lysis and need some stats for fuel injectors.</span></font></p> =0A=0A<p cl=
ass=3D"MsoNormal"><font size=3D"2" face=3D"Arial"><span style=3D"font-size:=
10.0pt;"> &nbsp;</span></font></p> =0A=0A<p class=3D"MsoNormal"><font size=
=3D"2" face=3D"Arial"><span style=3D"font-size:10.0pt;">What=0Adoes a typic=
al fuel injector used on a Renesis draw?&nbsp; I=E2=80=99m looking=0Afor 3 =
numbers:</span></font></p> =0A=0A<ol style=3D"margin-top:0in;" start=3D"1" =
type=3D"1">=0A <li class=3D"MsoNormal" style=3D""><font size=3D"2" face=3D"=
Arial"><span style=3D"font-size:10.0pt;">If I=E2=80=99ve got an injector on=
 the bench, what=0A     is the energized load?</span></font></li> =0A <li c=
lass=3D"MsoNormal" style=3D""><font size=3D"2" face=3D"Arial"><span style=
=3D"font-size:10.0pt;">When the injector is running in the engine, it is=0A=
     getting a PWM signal.&nbsp; What it the typical duty cycle at full=0A =
    throttle?&nbsp; This will help calculate average load over time.&nbsp; =
</span></font></li> =0A <li class=3D"MsoNormal" style=3D""><font size=3D"2"=
 face=3D"Arial"><span style=3D"font-size:10.0pt;">Or - is there some rule-o=
f-thumb number that is=0A     used for engine at full throttle?</span></fon=
t></li> =0A</ol>=0A=0A<p class=3D"MsoNormal"><font size=3D"2" face=3D"Arial=
"><span style=3D"font-size:10.0pt;"> &nbsp;</span></font></p> =0A=0A<p clas=
s=3D"MsoNormal"><font size=3D"2" face=3D"Arial"><span style=3D"font-size:10=
.0pt;">TIA</span></font></p> =0A=0A<p class=3D"MsoNormal"><font size=3D"2" =
face=3D"Arial"><span style=3D"font-size:10.0pt;"> &nbsp;</span></font></p> =
=0A=0A<p class=3D"MsoNormal"><font size=3D"2" face=3D"Arial"><span style=3D=
"font-size:10.0pt;">-Jeff</span></font></p> =0A=0A<p class=3D"MsoNormal"><f=
ont size=3D"2" face=3D"Arial"><span style=3D"font-size:10.0pt;"> &nbsp;</sp=
an></font></p> =0A=0A<p class=3D"MsoNormal"><font size=3D"2" face=3D"Arial"=
><span style=3D"font-size:10.0pt;"> &nbsp;</span></font></p> =0A=0A<p class=
=3D"MsoNormal"><font size=3D"2" face=3D"Arial"><span style=3D"font-size:10.=
0pt;">Jeff=0ALuckey</span></font></p> =0A=0A<p class=3D"MsoNormal"><font si=
ze=3D"2" face=3D"Arial"><span style=3D"font-size:10.0pt;">Direct=0AConnecti=
on Systems</span></font></p> =0A=0A<p class=3D"MsoNormal"><font size=3D"2" =
face=3D"Arial"><span style=3D"font-size:10.0pt;"><a rel=3D"nofollow" target=
=3D"_blank" href=3D"http://www.DCSIT.com">www.DCSIT.com</a></span></font></=
p> =0A=0A<p class=3D"MsoNormal"><font size=3D"2" face=3D"Arial"><span style=
=3D"font-size:10.0pt;">949-645-8832</span></font></p> =0A=0A<p class=3D"Mso=
Normal"><font size=3D"1" face=3D"Arial"><span style=3D"font-size:9.0pt;"> &=
nbsp;</span></font></p> =0A=0A<p class=3D"MsoNormal"><font size=3D"3" face=
=3D"Times New Roman"><span style=3D"=0Afont-size:12.0pt;font-family:&quot;T=
imes New Roman&quot;;"><br>=0A<br>=0A__________ Information from ESET NOD32=
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_<br>=0A<br>=0AThe message was checked by ESET NOD32 Antivirus.<br>=0A<br><=
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om</a></span></span></font></p> =0A=0A<p class=3D"MsoNormal"><font size=3D"=
3" face=3D"Times New Roman"><span style=3D"=0Afont-size:12.0pt;font-family:=
&quot;Times New Roman&quot;;"><br>=0A<br>=0A__________ Information from ESE=
T NOD32 Antivirus, version of virus signature=0Adatabase 3267 (20080714) __=
________<br>=0A<br>=0AThe message was checked by ESET NOD32 Antivirus.<br>=
=0A<br>=0A<a rel=3D"nofollow" target=3D"_blank" href=3D"http://www.eset.com=
">http://www.eset.com</a></span></font></p> =0A=0A</div>=0A=0A <br><br>____=
______ Information from ESET NOD32 Antivirus, version of virus signature da=
tabase 3267 (20080714) __________<br><br>The message was checked by ESET NO=
D32 Antivirus.<br><br><a rel=3D"nofollow" target=3D"_blank" href=3D"http://=
www.eset.com">http://www.eset.com</a><br> =0A=0A=0A</div></div><div style=
=3D"position:fixed"></div>=0A=0A=0A<!-- cg11.c3.mail.sp2.yahoo.com compress=
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