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From: "Ed Anderson" <eanderson@carolina.rr.com>
To: "'Rotary motors in aircraft'" <flyrotary@lancaironline.net>
Subject: RE: [FlyRotary] Re: Fuel Used was Alternator (Off topic)
Date: Thu, 19 Mar 2009 16:42:42 -0400
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Practice, Bob, Practice!  {:>)

Ed

Ed Anderson

Rv-6A N494BW Rotary Powered

Matthews, NC

eanderson@carolina.rr.com

http://www.andersonee.com

http://www.dmack.net/mazda/index.html

http://www.flyrotary.com/

http://members.cox.net/rogersda/rotary/configs.htm#N494BW

http://www.rotaryaviation.com/Rotorhead%20Truth.htm

-----Original Message-----
From: Rotary motors in aircraft [mailto:flyrotary@lancaironline.net] On
Behalf Of Bob White
Sent: Thursday, March 19, 2009 12:34 PM
To: Rotary motors in aircraft
Subject: [FlyRotary] Re: Fuel Used was Alternator (Off topic)

Ed,

All I want to know is how you did that conversion from gallon to shot
glass so quick.   ;)

Bob W.

On Thu, 19 Mar 2009 12:07:30 -0400
"Ed Anderson" <eanderson@carolina.rr.com> wrote:

> Ben, there is no question that anything consuming (using) electrical power
> is going to require that power to be produced (or stored) somewhere,
> sometime, somehow.  Whether you pull it from alternator or battery. 
> 
>  
> 
>  Since your objective (in this scenario) is to conserve fuel (rather than
> battery power),  the real question is how much fuel you will save vs. the
> utility of some electrical component you might decide to turn off.  For
> instance, you might save fuel by turning off your GPS and Radio - but if
the
> GPS is crucial to you taking the shortest, surest route to the nearest
> filling station - then keeping it ON is perhaps worth much more than the
few
> oz of fuel you might (I repeat might) save.
> 
>  
> 
> Lets look at some numbers.
> 
>  
> 
> At 12VDC 1 amp of current draw = 12 watts of power or much less than the
> ordinary 60 watt household light bulb, but probably enough for GPS and
Radio
> receiver.
> 
>  
> 
> 12 watts of power (assuming 100% efficiency which you, of course, will NOT
> get) would require  0.01609227 HP.  Lets round it up and say 0.0170 HP
> 
>  
> 
> A frequently used power equation for the rotary is HP = lbm fuel/0.55.  So
> reworking the equation slightly  we get Fuel = HP*0.55  = 0.0170*0.55 =
> 0.00935 lb of fuel/min
> 
>  
> 
> So lets say you need to fly 30 minutes to get to the nearest airport.
> 0.00935*30 = 0.2804 lb fuel required to power the 12 watt component for 30
> minutes.
> 
>  
> 
> Since there is approx 6 lbs of fuel per gallon, you would use 0.2805 / 6 =
> 0.0467 gallons of fuel or approx 6 oz of fuel (three jigger size shot
> classes worth - you might need that after this flight {:>)).
> 
>  
> 
> At a throttled back cruise rate of  4 gallon per hour of fuel burn (if you
> can lean it back that far and stay airborne and get good mileage) you
would
> save enough to keep the engine running  
> 
>  
> 
> 4 gallons/hr =  512 oz (approx) /hr   , So 6 oz = 6/512 *60 minutes =
0.703
> minute or  42 seconds longer of flight (very approximate)
> 
>  
> 
>  
> 
> So the bottom line is the 42 seconds of flight more valuable than having
> your GPS and radio - it could be if that made the difference between
making
> the airport or not (but you don't know that bit of important information),
> but then if you lose the most direct route to your airfield because you
> turned off your GPS, you could easily use much more fuel just doing a 360
> turn or two.
> 
>  
> 
> On the other hand, if your battery was fully charged  then you might not
> draw the battery down enough to trigger the voltage regulator to send
power
> to it, in which case you may not save any fuel (over this short time
period)
> by giving up your 12 watt GPS /Radio.
> 
>  
> 
>  
> 
> Hypothetical scenarios are fun to discuss and give you a ball park
> quantification of factors that might turn out to be useful in decision
> making (should you every face the situation), but my view is it is seldom
> you find yourself in a real world situation that other factors don't screw
> up the decision that the hypothetical scenario would call for {:>)
> 
>  
> 
>  
> 
> Just my 0.02 worth
> 
>  
> 
> Ed
> 
>  
> 
> Ed Anderson
> 
> Rv-6A N494BW Rotary Powered
> 
> Matthews, NC
> 
> eanderson@carolina.rr.com
> 
>  <http://www.andersonee.com> http://www.andersonee.com
> 
>  <http://www.dmack.net/mazda/index.html>
> http://www.dmack.net/mazda/index.html
> 
> http://www.flyrotary.com/
> 
>  <http://members.cox.net/rogersda/rotary/configs.htm>
> http://members.cox.net/rogersda/rotary/configs.htm#N494BW
> 
> http://www.rotaryaviation.com/Rotorhead%20Truth.htm
> <http://www.dmack.net/mazda/index.html> 
> 
>   _____  
> 
> From: Rotary motors in aircraft [mailto:flyrotary@lancaironline.net] On
> Behalf Of Ben Baltrusaitis
> Sent: Thursday, March 19, 2009 9:09 AM
> To: Rotary motors in aircraft
> Subject: [FlyRotary] Alternator (Off topic)
> 
>  
> 
> Since it's quiet:
> 
>  
> 
> When I was a kid a guy at the parts store demonstrated to my Dad that when
> electrical power was needed, a generator put a load on the engine. After
> that, my Dad was careful not to run lights, radio, heater fan, or other
> non-essentials when he was trying to get good gas mileage.
> 
>  
> 
> I have continued that tradition, however, I have seen it stated that
> electrical draw on an alternator doesn't increase the mechanical load.
> 
>  
> 
> When low on fuel will it help to turn off electrical components not needed
> for flight?
> 
>  
> 
> Is it true of an alternator; an electrical power demand doesn't cause an
> increased mechanical load?
> 
>  
> 
> Or, does keeping headlights on during the day decrease gas mileage?
> 
>  
> 
> Thanks!
> 
> Ben
> 
> 
> 
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-- 
N93BD - Rotary Powered BD-4 - http://www.bob-white.com
3.8 Hours Total Time and holding
Cables for your rotary installation - http://roblinstores.com/

--
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