Today's Message Index:
----------------------
1. 05:41 AM - Re: Jabiru Voltage Regulator Low Voltage Light (messydeer)
2. 08:02 AM - Re: Re: Jabiru Voltage Regulator Low Voltage Light (Michael Welch)
3. 09:15 AM - Reduce PS voltage (jonlaury)
4. 10:59 AM - Re: Reduce PS voltage (Ken)
5. 02:26 PM - Re: battery charger (James Robinson)
6. 05:13 PM - Re: Reduce PS voltage (Richard E. Tasker)
7. 07:20 PM - Re: Reduce PS voltage (Ken)
Message 1
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| Subject: | Re: Jabiru Voltage Regulator Low Voltage Light |
Thanks, Mike and Bob.
Yes, I'll get downward pressure on the tail next time. I've driven a couple pieces
of ~1" pipe ~18" into the ground angled back and away from each other, about
2' apart. I'll tie a rope near the ground on one, loop it up 6" to go through
the bottom end of the tailcone, then back down to the other pipe. I'm tempted
also to put a 50# weight on the end. Right now, there's not much pressure on
the tailspring, maybe 20-30 lbs. The wheels being chocked also caused it to
nose over, but I think I'll keep them.
--------
Dan
Read this topic online here:
http://forums.matronics.com/viewtopic.php?p=355719#355719
Message 2
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| Subject: | Re: Jabiru Voltage Regulator Low Voltage Light |
Hi Dan,
It's an excellent idea to hold the tail down better next time. Wheel chocks
and ropes straight back to trees
are a great way to keep the plane from rolling forward, but they do virtually nothing
to keep the propeller from
driving the nose into the dirt.
A 50# weight tied close to the tail will go a loooong way to keep the plane from
catastrophic disaster. The secure
steel pipes are an excellent idea, too......and also the minimum you should consider.
The pipes and the weight will
guarantee you will not have that situation reoccur. I'm glad to know you are
fixing things.
Mike Welch
On Oct 22, 2011, at 7:29 AM, messydeer wrote:
>
> Thanks, Mike and Bob.
>
> Yes, I'll get downward pressure on the tail next time. I've driven a couple pieces
of ~1" pipe ~18" into the ground angled back and away from each other, about
2' apart. I'll tie a rope near the ground on one, loop it up 6" to go through
the bottom end of the tailcone, then back down to the other pipe. I'm tempted
also to put a 50# weight on the end. Right now, there's not much pressure
on the tailspring, maybe 20-30 lbs. The wheels being chocked also caused it to
nose over, but I think I'll keep them.
>
> --------
> Dan
>
>
>
>
> Read this topic online here:
>
> http://forums.matronics.com/viewtopic.php?p=355719#355719
>
>
>
>
>
>
>
>
>
>
>
Message 3
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| Subject: | Reduce PS voltage |
I'm trying to reform the caps in an old unused Whelen strobe PS by using an independent
PS.
A Whelen tech suggested powering up the Whelen PS at 6vdc to see if it works. And
if it does, then power up to 12v
Whelen's white paper says use 9v for 15 min, then 12v.
Other people have reformed in steps, 6v, 9v, 12v.
How do I step down the voltage of my independent 13.5 PS?
The Whelen PS says it uses 7a. I tried using 4 'C' cells in series but the voltage
dropped to mV, so I'm deducing that I need an independent supply capable of
more amperage. At 6v, will I be trying to dissipate 7.5v x 7a = 42W? That's
a lot of heat if I just add a big honkin' resistor to the 13.5 v supply. Don't
know if there are resistors that big. I think there must be a more sophisticated
way to do this.
Thanks,
John
Read this topic online here:
http://forums.matronics.com/viewtopic.php?p=355729#355729
Message 4
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| Subject: | Re: Reduce PS voltage |
I think I used a 12 volt headlight bulb as a "resistor" when I did this
but too long ago to be sure. IIRC the average current is considerably
less than 7 amps. Suspect I would have paralleled a second bulb for the
intermediate voltage. I always keep a spare headlight bulb or two in
stock. Other things that you can play with include resistance
appliances. A 1500 watt kettle would be around 1500/120v = 12.5 ohms.
Two in parallel would be 6.25 ohms.
Ken
On 22/10/2011 11:58 AM, jonlaury wrote:
> --> AeroElectric-List message posted by:
> "jonlaury"<jonlaury@impulse.net>
>
> I'm trying to reform the caps in an old unused Whelen strobe PS by
> using an independent PS. A Whelen tech suggested powering up the
> Whelen PS at 6vdc to see if it works. And if it does, then power up
> to 12v Whelen's white paper says use 9v for 15 min, then 12v.
>
> Other people have reformed in steps, 6v, 9v, 12v.
>
> How do I step down the voltage of my independent 13.5 PS?
>
> The Whelen PS says it uses 7a. I tried using 4 'C' cells in series
> but the voltage dropped to mV, so I'm deducing that I need an
> independent supply capable of more amperage. At 6v, will I be trying
> to dissipate 7.5v x 7a = 42W? That's a lot of heat if I just add a
> big honkin' resistor to the 13.5 v supply. Don't know if there are
> resistors that big. I think there must be a more sophisticated way
> to do this.
>
> Thanks, John
>
>
Message 5
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| Subject: | Re: battery charger |
Hi Bob=0AI went to WalMart but none of the chargers had model #s- Any oth
er description ? I have a battery tender now , does the Schumacher unit off
er more options?=0A=0AJim=0A=0A-=0AJames Robinson=0AGlasair lll N79R=0AS
panish Fork UT U77=0A=0A=0A________________________________=0AFrom: "Rober
t L. Nuckolls, III" <nuckolls.bob@aeroelectric.com>=0ATo: aeroelectric-list
@matronics.com=0ASent: Wednesday, August 17, 2011 9:55 AM=0ASubject: Re: Ae
roElectric-List: battery charger=0A=0A--> AeroElectric-List message posted
by: "Robert L. Nuckolls, III" <nuckolls.bob@aeroelectric.com>=0A=0AAt 12:44
AM 8/17/2011, you wrote:=0A> Hi Bob=0A> I was looking for the info on the
battery chargers and I must not have saved it.- It seems you recommended
one available from WalMart that was a good value=0A> Jim=0A> =0A=0A- I t
hink you're remembering the Schumacher=0A- model 1562.=0A=0A=0A=0A- Bo
======================
Message 6
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| Subject: | Re: Reduce PS voltage |
Good ideas, but your calculations for a 1500 watt kettle are wrong. You have calculated
the current consumption (Current=Watts/Voltage). To calculate ohms
use: watts = (voltage * voltage) /
resistance or more specifically resistance = (voltage * voltage) / watts. Or in
this case (120*120)/1500 = 9.6 ohms. It is probably somewhat less than this
since it won't heat to the normal
operating temperature and the resistance goes up as it heats.
Dick Tasker
Ken wrote:
>
> I think I used a 12 volt headlight bulb as a "resistor" when I did this
> but too long ago to be sure. IIRC the average current is considerably less than
7 amps. Suspect I would have paralleled a second bulb for the
> intermediate voltage. I always keep a spare headlight bulb or two in stock. Other
things that you can play with include resistance appliances. A 1500 watt
kettle would be around 1500/120v = 12.5
> ohms. Two in parallel would be 6.25 ohms.
> Ken
>
> On 22/10/2011 11:58 AM, jonlaury wrote:
>> --> AeroElectric-List message posted by:
>> "jonlaury"<jonlaury@impulse.net>
>>
>> I'm trying to reform the caps in an old unused Whelen strobe PS by
>> using an independent PS. A Whelen tech suggested powering up the
>> Whelen PS at 6vdc to see if it works. And if it does, then power up
>> to 12v Whelen's white paper says use 9v for 15 min, then 12v.
>>
>> Other people have reformed in steps, 6v, 9v, 12v.
>>
>> How do I step down the voltage of my independent 13.5 PS?
>>
>> The Whelen PS says it uses 7a. I tried using 4 'C' cells in series
>> but the voltage dropped to mV, so I'm deducing that I need an
>> independent supply capable of more amperage. At 6v, will I be trying
>> to dissipate 7.5v x 7a = 42W? That's a lot of heat if I just add a
>> big honkin' resistor to the 13.5 v supply. Don't know if there are
>> resistors that big. I think there must be a more sophisticated way
>> to do this.
>>
>> Thanks, John
>>
>>
>
>
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Message 7
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| Subject: | Re: Reduce PS voltage |
Yes thanks - I should read before sending.
1500 watts/120volts = 12.5 AMPS
120volts/12.5 amps=9.6 ohms
Ken
On 22/10/2011 8:08 PM, Richard E. Tasker wrote:
> <retasker@optonline.net>
>
> Good ideas, but your calculations for a 1500 watt kettle are wrong. You
> have calculated the current consumption (Current=Watts/Voltage). To
> calculate ohms use: watts = (voltage * voltage) / resistance or more
> specifically resistance = (voltage * voltage) / watts. Or in this case
> (120*120)/1500 = 9.6 ohms. It is probably somewhat less than this since
> it won't heat to the normal operating temperature and the resistance
> goes up as it heats.
>
> Dick Tasker
>
> Ken wrote:
>>
>> I think I used a 12 volt headlight bulb as a "resistor" when I did this
>> but too long ago to be sure. IIRC the average current is considerably
>> less than 7 amps. Suspect I would have paralleled a second bulb for the
>> intermediate voltage. I always keep a spare headlight bulb or two in
>> stock. Other things that you can play with include resistance
>> appliances. A 1500 watt kettle would be around 1500/120v = 12.5 ohms.
>> Two in parallel would be 6.25 ohms.
>> Ken
>>
>> On 22/10/2011 11:58 AM, jonlaury wrote:
>>> "jonlaury"<jonlaury@impulse.net>
>>>
>>> I'm trying to reform the caps in an old unused Whelen strobe PS by
>>> using an independent PS. A Whelen tech suggested powering up the
>>> Whelen PS at 6vdc to see if it works. And if it does, then power up
>>> to 12v Whelen's white paper says use 9v for 15 min, then 12v.
>>>
>>> Other people have reformed in steps, 6v, 9v, 12v.
>>>
>>> How do I step down the voltage of my independent 13.5 PS?
>>>
>>> The Whelen PS says it uses 7a. I tried using 4 'C' cells in series
>>> but the voltage dropped to mV, so I'm deducing that I need an
>>> independent supply capable of more amperage. At 6v, will I be trying
>>> to dissipate 7.5v x 7a = 42W? That's a lot of heat if I just add a
>>> big honkin' resistor to the 13.5 v supply. Don't know if there are
>>> resistors that big. I think there must be a more sophisticated way
>>> to do this.
>>>
>>> Thanks, John
>>>
>>>
>>
>>
>>
>>
>>
>
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