---------------------------------------------------------- AeroElectric-List Digest Archive --- Total Messages Posted Sat 10/22/11: 7 ---------------------------------------------------------- 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 _____________________________________ Time: 05:41:58 AM PST US Subject: AeroElectric-List: Re: Jabiru Voltage Regulator Low Voltage Light From: "messydeer" 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 _____________________________________ Time: 08:02:32 AM PST US From: Michael Welch Subject: Re: AeroElectric-List: 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 _____________________________________ Time: 09:15:47 AM PST US Subject: AeroElectric-List: Reduce PS voltage From: "jonlaury" 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 _____________________________________ Time: 10:59:43 AM PST US From: Ken Subject: Re: AeroElectric-List: 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" > > 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 _____________________________________ Time: 02:26:05 PM PST US From: James Robinson Subject: Re: AeroElectric-List: 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" =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" =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 _____________________________________ Time: 05:13:34 PM PST US From: "Richard E. Tasker" Subject: Re: AeroElectric-List: 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" >> >> 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 >> >> > > -- Please Note: No trees were destroyed in the sending of this message. We do concede, however, that a significant number of electrons may have been temporarily inconvenienced. -- ________________________________ Message 7 _____________________________________ Time: 07:20:29 PM PST US From: Ken Subject: Re: AeroElectric-List: 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: > > > 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" >>> >>> 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 >>> >>> >> >> >> >> >> > ------------------------------------------------------------------------------------- Other Matronics Email List Services ------------------------------------------------------------------------------------- Post A New Message aeroelectric-list@matronics.com UN/SUBSCRIBE http://www.matronics.com/subscription List FAQ http://www.matronics.com/FAQ/AeroElectric-List.htm Web Forum Interface To Lists http://forums.matronics.com Matronics List Wiki http://wiki.matronics.com Full Archive Search Engine http://www.matronics.com/search 7-Day List Browse http://www.matronics.com/browse/aeroelectric-list Browse Digests http://www.matronics.com/digest/aeroelectric-list Browse Other Lists http://www.matronics.com/browse Live Online Chat! http://www.matronics.com/chat Archive Downloading http://www.matronics.com/archives Photo Share http://www.matronics.com/photoshare Other Email Lists http://www.matronics.com/emaillists Contributions http://www.matronics.com/contribution ------------------------------------------------------------------------------------- These Email List Services are sponsored solely by Matronics and through the generous Contributions of its members.