Have you noticed that all plane crashes are as a result of pilot error?  No other cause…only other factors.

 

I read one the other day where the mechanic failed to install a cotter pin on the throttle body linkage and the engine quit on takeoff…The cause of the crash was the pilot failing to control the plane and impacting terrain, a FACTOR was the missing cotter pin!??

 

Don’t blame me, this is what happens when someone says the list is quiet. 

 

Bill B

From: Rotary motors in aircraft [mailto:flyrotary@lancaironline.net] On Behalf Of Ed Anderson
Sent: Thursday, March 19, 2009 12:08 PM
To: Rotary motors in aircraft
Subject: [FlyRotary] Fuel Used was Alternator (Off topic)

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

 

 

__________ Information from ESET NOD32 Antivirus, version of virus signature database 3267 (20080714) __________

The message was checked by ESET NOD32 Antivirus.

http://www.eset.com