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[h.a.c. van asten]

LJ.Dnld . The fist leg of 874-900 mls asked 10% more fuel burnt whereas a 44 mintes delay on ETA-Nukumanu was incurred . Recomputation (1996) showed that @ 1912 GMT the reserves were sufficient for 1h05m , the 100 oct special avgas included . The hourly mean gas consumption was (1,100 US gals.) / 20h17m = 54 gls. These figures when fed into the entire , 2729 mls  , flight´s performance formula give good compliance with 20 mph averaged headwinds and groundspeed 134-136 mph.  154 mph /54 gph ( 1 - 20 mph /154 mph) = 2.48 mpg , 2 mls taxi-and-run distance included. [154 mph = Ind.Air Speed , 54 gph = US gls/hr , 20 mph = av.equiv. headwinds]. The 1996 inquiry needs updating due to new vistas but however , figures given will not essentially decline. Initial fuel loaded was 1,050 gls 80 oct + 50 gls 100 oct (remaining after Lae take off).

Sry the 100 oct after takeoff Lae was 22 US , not 100 . H

[Martin X. Moleski, SJ]

... As to mixture, we'll never know for sure how good her technique was, but that too should have been elementary enough with the aid of the Cambridge analyzer on board. 

She seems to have spent two days in Bandoeng (June 25-26) just getting the exhaust analyzer working.  That suggests to me that she was using it en route, knew when it wasn't working, and wanted to have it in good order for the big over-water legs.

Even failing that instrument, leaning to point of misfire and enrichening to smooth engine again and watching cylinder head temps is almost instinctive. 

Agreed.  She was, in many respects, an airhead (pun intended; you may laugh now), but she couldn't have made as many long-distance flights as she did without understanding the basics of fuel and power management.

[Mona Kendrick]

she couldn't have made as many long-distance flights as she did without understanding the basics of fuel and power management.
[/quote]

     I would agree.  Bernt Balchen mentions in Come North with Me having calculated a throttle curve for use on her Atlantic solo flight, so we know she was familiar with the concept of progressive power reduction at least as early as 1932.  That she was already quite familiar with mixture control by then is implied by her use of mixture settings as a sort of supplemental source of altitude information after the altimeter failed on that flight: taking note of how far back the mixture could be leaned gave a very rough indication of current altitude.

LTM,
Mona

[h.a.c. van asten]

I forgot the articles on fuel etc. are in Dutch , I will translate them ; article on fuel load (1996 matter of dispute ) first (from "Newsletter" RDAFM , nr.58 , July 1996) and post when finished  , soon as possible .

[h.a.c. van asten]

Yes , but a headwind or equivalent headwind blowing 20 mph sets any specific flown 10E distance per gallon gasoline (rich or lean)  back with 13 percent , if your emergency reserve was 10% , you may get 3% short .

[h.a.c. van asten]

She herself said the 1912 87 oc store to be for 1 / 2 hour , together giving 1h05m with the 100 oc remaining . Normally a 10E could fly at specific 150 mph / 38 gph = 3.94 mls/ gallon . A 20 mph headwind aloft sets this back to 3.43 mpg , or 13% less . On the Lae-Nukumanu leg with 25 mph eastern the mpg figure was set back with 17% . Try : the formula is  V ground p.hr / fuel expenditure p.hr = V true airspeed / fuel expenditure p.hr  X  [ (1 - V wind / V true airspeed)] . Like by cars on the highway , the air resistance is predominant over other resistances in flat country . Doubling your speed asks for  2 to power 3 = 8 times more labour . Maintaining your speed with head-on wind  asks for (V-own + V-wind) to power 3 additional labour  , at any engine power setting .

[Irvine John Donald]

So if there is somewhere between 50 and 100 gallons remaining once landed on Gardner is this enough for several days of running up the engine to charge radio batteries?  I think i read something that said the engine had to be run at a setting higher than an idle to run the radio while charging the battery. Does the remaining fuel burned match the number of days of post loss radio messages?

If yes then this lends further credibility to the hypothesis of Gardner being the likely final landing spot. If no then is it because they would have run out sooner than the messages ended, thus casting doubt on the authenticity of the last few post error messages. Or if they had gas for more messages then why did they stop transmitting?  Electra went over reef edge into deeper water? Or castaways passed away? Or...

Does the fuel consumption fit the hypothesis?  Not to the last drop but at least within the scenario and knowing we dont know the exact burn rate.

[h.a.c. van asten]

At 1912 GMT about 22 gls 100-oc remained from 50 . The store than was  23 gls 87-oc + 22 gls 100-oc = 45 US . @ 41 gph in low dense air the remaining flight time was 1h05m - 1h06 m . Add to 1912 = 2017 GMT . Before running dry however , carbureters deliver up and down supply . The remaining 22-oc gasoline was used from 1845 GMT , switching to container was possibly the reason why no radio call was given at cue ; it being also possible that crew was busy trying to receive bearing result that was asked for @ 1815 GMT . Air resistance , btw , is for land vehicles same as for aircraft : the labour needed is  Pa = 1/2 R . Cw . A . V^3 with Pa = air resistance x way ; R = air density ; Cw = resistance coefficient ; V = speed . If measured in S.I. , outcome is in dimension Newton x m/s = Watts . The Johnson reports have for the inquiry not been used , other than for gph figures predicted ; remaining figures come from meteo versus true weather & radio calls . I am busy translating the 1996 articles , first on fuel loading @ Lae . H.

[Martin X. Moleski, SJ]

So if there is somewhere between 50 and 100 gallons remaining once landed on Gardner is this enough for several days of running up the engine to charge radio batteries?  I think i read something that said the engine had to be run at a setting higher than an idle to run the radio while charging the battery. Does the remaining fuel burned match the number of days of post loss radio messages?

Here is an interim report from Bob Brandenburg on that score.

Without doing the math, I think the answer is "yes."

[Chuck Varney]

Here is an interim report from Bob Brandenburg on that score.

Marty,

Bob might want to correct the opening words of the seventh paragraph: "The transmitter high-voltage power supply was a dynamotor that drew 65 amps at 12 volts. . ." As he correctly indicates further on, it was the transmitter that drew approximately 65 A at 12 V under modulated conditions--not the dynamotor alone.

Chuck

[Martin X. Moleski, SJ]

Here is an interim report from Bob Brandenburg on that score.

Bob might want to correct the opening words of the seventh paragraph: "The transmitter high-voltage power supply was a dynamotor that drew 65 amps at 12 volts. . ." As he correctly indicates further on, it was the transmitter that drew approximately 65 A at 12 V under modulated conditions--not the dynamotor alone.

Yes, I see.  For the moment, I've amended the sentence to: "The transmitter drew 65 amps at 12 volts from a dynamotor, and thus could function at virtually full output even with the battery approaching zero charge."

[Chuck Varney]

For the moment, I've amended the sentence to: "The transmitter drew 65 amps at 12 volts from a dynamotor, and thus could function at virtually full output even with the battery approaching zero charge."

Marty,

The transmitter drew 65 A at 12 V from a battery source, not from the dynamotor. Deleting "from a dynamotor" would fix your amended sentence.

Chuck

[Chuck Varney]

Not to put too fine a point on it, but we should be careful about how we treat Brandenburg's statement - the transmitter only draws from the battery if you consider the dynamotor to be one and the same with the transmitter device.  It was not. 

Jeff,

Let's do put a finer point on it.

You wrote: ". . .the transmitter only draws from the battery if you consider the dynamotor to be one and the same with the transmitter device." That is not a true statement.

The dynamotor was not the only 12-volt transmitter-related load. The tube filaments were heated directly by 12 volts (~6 A standby, 10-11 A transmit). The oscillator crystal heater option, if used, operated on 12 volts. I think the control circuits were 12-volt, as well. It  was the total of these loads, when transmitting, that equaled approximately 65 A.


Chuck

[Martin X. Moleski, SJ]

For the moment, I've amended the sentence to: "The transmitter drew 65 amps at 12 volts from a dynamotor, and thus could function at virtually full output even with the battery approaching zero charge."

The transmitter drew 65 A at 12 V from a battery source, not from the dynamotor. Deleting "from a dynamotor" would fix your amended sentence.

I thought the dynamotor was an essential part of the operation--isn't the current going through it from the batteries through the dynamotor to the TX?

After further review, the offending sentence now begins, "The transmission system drew 65 amps at 12 volts, ..."  I've dropped a box in the middle of the passage that gives more details about the dynamotor, the heating circuits, and the different operating modes.

[Chuck Varney]

I thought the dynamotor was an essential part of the operation--isn't the current going through it from the batteries through the dynamotor to the TX?

Marty,

You're correct. The dynamotor was an essential part of the operation--and the first three sentences in your added box summarize the situation nicely. I've apparently done a grand job of unintentional obfuscation here, but all I wanted to point out was that it was all the 12-volt transmitter-related loads--including the dynamotor--that summed to 65 A, not just the dynamotor, as Bob stated it.

It's with great trepidation that I make another comment to the same writeup. In the sentence before the one under discussion, Bob wrote "Aircraft lead-acid battery cells of the period had a specific gravity range of 1300 at full charge to 1110 at fully discharged, with corresponding voltage range of 12.86 volts to 11.81 volts." The specific gravities should be 1.300 and 1.100, not 1300 and 1100.

Chuck