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h.a.c. van asten

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« on: August 14, 2011, 04:05:21 AM »

Apply some Newtonian Aero Dynamics : The lift needed for an A/c in horizontal flight is from CL . 1/2 . r . S . Vt^2 = m. g . in th dimension Newton (N) . For NR 16020 @ 8,000 ft with r = 0.964 kg/m^3 , S = 42.6 m^2 surface wing , Vt = 69 m/s TAS , Vw = 8 m/s headwind , average mass = 4,932 kg in flight we find : m . g  = 48,383 N with CL = 0.459 lift coefficient . Reasonable glide ratio CL/CD Lockheed 10  is 10 (report Lockheed no.487) . Thence : 48,383 N / 10 =  4,843 is the propeller draft needed . The mechanical propulsive efficiency  ep was 21.6% (0.76 for propellers , Report 487 ; & reasonable 0.28 for the engines) so that for a range R follows : E = (m . g . R) ep^-1 , or by insertion : 4.63(4,838 N x 2,628 mls x 1,6092 m/mi x 1,000 m/km) = 9.47 x 10^10 kg . m^2 . s^-2 in the energy dimension  J  for Joule .  Divide by 47 x 10^6 J / kg gasoline chemical energy and find  2,015 kg , 4,438 lbs , or 740 US gallons for 2,528 mls Lae to Howland , zero wind , duration 19h12m , offset included . The power needed to pull 48,393 N is  CL . 1/2 . r . S . Vt^3  in zero wind , to be transposed to CL . 1/2 . r . S . (Vt + Vw)^3 in the event of headwind speed Vw . (Vt + Vw)^3 = 1.39 Vt^3 . Note that any acceleration factor against resistance translates to the  factor in 3rd power to energy needed  (2 x as fast = 8 x greater energy) . Hence : needed for 2,628 mls in headwind 18 mph , 8 m/s , Beaufort 6 is : 740 galls x 1.39 = 1,028 galls . Remaining @ GMT 1912  22 galls 87 oc & max. 25 galls 100 oc fuel from 1,100 initial store . The specific range was 2,628 mls / 1,028 galls =  2.56 mph , so that by extrapolation follows that after GMT 1912  a maximum distance  47 x 2.56 mls = 120 mls  could be additionally flown . This range is from the Howland region too small for reaching any other in the Pacific land point than Howland itself , or Baker .

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Gary LaPook

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« Reply #1 on: August 14, 2011, 10:28:05 AM »

------------------------------------------------------------

Also see excerpts from Boilerman 3 & 2, the U.S. Navy training manual for boilermen 3rd and 2nd class which is attached. BTW, nowhere in this manual or in Knight's Modern Seamanship is there any mention of the horribles that inhabit Brandenburg's fertile imagination.

- Uneven heating among the tubes, etc. create all kinds of stress; relative cooling in one section / component and extreme heat in the next creates the potential for what amounts to a bomb - it's basics 101 in boiler-world.  Point is to do what was described is counter to good practices and it was not likely what really happened.  In any case, AE certainly never reported finding any smoke that we're aware of... the rest may be - speculation.  I appreciate your insight, but also happen to be long familiar with steam engines of many sorts. - Jeff
-------------------------------------------------------

Wow Jeff, I just realized that you must be very, very old! I think the last time an A&P worked on a steam powered airplane it was Langley's in 1903.

gl
« Last Edit: August 16, 2011, 07:39:47 AM by Martin X. Moleski, SJ »
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Harry Howe, Jr.

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• Nuclear Physicist(Ret) Pilot(Ret) Scuba(Ret)
« Reply #2 on: August 14, 2011, 02:30:40 PM »

H.A.C
And according to Newtonian Aerodynamics, a Bee can'fly!  Tell that to them the next time you disturb a hive full of them.

Do it the simple way:
The Lockheed report gives the range of the 10E as between 4100 and 4500 miles on 1200 gallons of fuel.  Use the midpoint 4300 miles and divide by 1200 and get 3.5833 miles per gallon.
The Lae to Howland distance is 2556 miles, divided by 3.5833 eequals 713.3 gallons which when subtracted from 1100 gallons leaves386.7gallons, multiplied by 3.5833 m/g gives 1386 miles, enough to go to Gardner and back 3 times.
Which of course we gcould have gotten by subtracting the distance Lae to Howland, 2556 or so from 4300 to get 1744.
Conclusion:  They had more than enough feul to fly to Gardner, land, and run the starboard engine to charge the batteries to operate the radio to send out distress calls.
No Worries Mates
LTM   Harry (TIGHAR #3244R)

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Harry Howe, Jr.

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• Posts: 576
• Nuclear Physicist(Ret) Pilot(Ret) Scuba(Ret)
« Reply #3 on: August 14, 2011, 03:18:26 PM »

Obviously I had a brain twitch in my last post

It should be BEES can't fly

and the 3 was wrong, they could only have flown from Howland and back and then back to Gardner. 1-1/2 round trips, not 3.
No Worries Mates
LTM   Harry (TIGHAR #3244R)

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Monty Fowler

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« Reply #4 on: August 14, 2011, 03:39:18 PM »

Maybe bees can't fly, but they can sure sting the bejesus outta you. I'm sure Mr. van Asten has a mathematical equation about 50 pages long for that, as well.

LTM,
TIGAR No. 2189CE, wishing the Discovery Channel would rebroadcast Finding Amelia in the mainstream instead of the backwaters.
Ex-TIGHAR member No. 2189 E C R SP, 1998-2016

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h.a.c. van asten

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• Posts: 322
« Reply #5 on: August 15, 2011, 01:32:07 AM »

A very good assessment indeed . Do not however , forget the continuous headwind pressure : your outcome must be for required energy multiplied by (Vg + Vw)^3 / Vg^3 = 1.44 for this case . I inserted an error figure iin the comment  (Vg was exchanged for Vt) , the reciprocal of the propulsive efficiency  1/ep = 4.46 giving 9.14 x 10^10 J , and 4,284 lbs fuel for 714 (your figure) galls . This figure to multiply by 1.44 gives 1,028 galls . The same figure appears when calculated by starting @ the s.c. wind regression factor . So , without  the headwinds asking for (1.13)^3 = 1.44 more energy w.r.t. to zero wind or moderate wind resistance , the GMT 1912 reserve would have been (1,100 - 25) - 714 = 361 galls maximum , for 924 mls flight (there and back Gardner) as you suppose . When different calculation algorithms lead to a same outcome , it is always a signal that computations are correct . Probably , in Report 487 was reckoned for incidental , moderate wind fields , not with actual Trade Winds averaging , head  or cross , an equivalent 18 mph for the endurance . The fuel consumption per hour was btw normally 310 lbs/hr (letter Putnam to De Sibour , Feb 13 , 1937) . The actual trip asked 321.3 galls/hr , 3.6% more only . In 2,628 mls is 32 mls offset detour included , leaving 40 mls detour w.r.t. 2,556 , given the navigation model I worked with for the EJN-2008 article .
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Gary LaPook

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« Reply #6 on: August 15, 2011, 02:20:26 AM »

H.A.C
And according to Newtonian Aerodynamics, a Bee can'fly!  Tell that to them the next time you disturb a hive full of them.

Do it the simple way:
The Lockheed report gives the range of the 10E as between 4100 and 4500 miles on 1200 gallons of fuel.  Use the midpoint 4300 miles and divide by 1200 and get 3.5833 miles per gallon.
The Lae to Howland distance is 2556 miles, divided by 3.5833 eequals 713.3 gallons which when subtracted from 1100 gallons leaves386.7gallons, multiplied by 3.5833 m/g gives 1386 miles, enough to go to Gardner and back 3 times.
Which of course we gcould have gotten by subtracting the distance Lae to Howland, 2556 or so from 4300 to get 1744.
Conclusion:  They had more than enough feul to fly to Gardner, land, and run the starboard engine to charge the batteries to operate the radio to send out distress calls.

------------------------------------------------------------------------

Further fuel analysis.

Looking at figure VI on page 12 of Lockheed Report 487 we can see that 1200 gallons would provide
a maximum range of 4060 statute air miles;
1150 gallons will take you 3940;
1100 gallons will take you 3800 and
1050 gallons will take you 3645 statue air miles.

The minimum fuel needed to go the 2556 statute miles distance to Howland in still air is 680 gallons
meaning that if the plane took off at the light gross weight of 13,324 with just 680 gallons it could fly
that many miles. Taking off with more fuel weight causes the plane to burn more fuel en route so more
fuel will be consumed on the way to Howland when starting with a greater fuel load.

Boswell did an analysis and, starting with 1150 gallons, calculated the maximum range of 4160 statute
air miles, 220 miles more than report 487. Analyzing his report, you find his calculations shows the
plane burned 871 gallons in 18:32 minutes needed to fly the 2556 miles to Howland leaving 279 gallons
remaining and a further range of 1604 statute air miles.

There is a third way to do this calculation, using the Breguet formula, which  needs very few inputs and
produces amazingly accurate results.

Range (statute air miles) = 863.5 times L/D (max) times propeller efficiency divided by specific fuel
consumption times the Log of the starting weight divided by the finishing weight. See attached pages 186 and 187
from “Airplane Performance Stability and Control" by Perkins & Hage.

formula 4-40 on page 186

R= 863.5 L/D n/c LOG (W0/W1)

To use this formula we only need to find coefficient of lift / coefficient of drag that is maximum, or
maximum Lift over Drag,

L/D (max)

This information is found on page 30 of the Lockheed report. (There are a number of typos on this
page that I have marked on the attached page.) Looking in the first column under “CL” for 16,500 #
we find the coefficient of lift at 150 mph is .63. We find the coefficient of drag (CD) for the same speed
and weight (after correcting the typo) is .053. Dividing CL by CD we find CL/CD = 11.89. We can do
the same calculation with other pairs of values and we will find that 150 mph produces the maximum
L/D at 16500#. We can also calculate the maximum at the other listed weights and we will find that the
speeds for maximum range varies with the square root of the weight ratio just as the laws of
aerodynamics predicts and the power for these speeds also change with the 3/2 power of the weight
ratios just as predicted by the same laws.

Next we find propeller efficiency on page 32. It varies a bit but we can choose .75 Similarly we find
SFC on page 34 of .46 pounds per hour per horsepower. Substituting these values into the formula we
end up with a constant times the LOG of the weight ratio.

.75/.46 x 11.89 x 863.5 = 16739 so range = 16739 times LOG of the weight ratio.

The finish weight is 9300# per the Lockheed report. Starting with 1200 gallons, gross weight 16500 the
calculated range is 4154 statute air miles.
1150 gallons produces a range of 4021, 1465 remaining after flying the distance to Howland;
1100 gallons ...............................3886, 1330....;
1050............................................3748, 1192.....

Further manipulation of the formula allows us to calculate the weight after flying 2556 statute air miles.
Starting with 1200 gallons the weight after 2556 miles will be 11596 meaning 817 gallons used;
Starting with 1150.................................................................11384..............803....................;
....................1100..................................................................11174..............788....................;
.....................1050................................................................10963................773 gallons used.

Also it would take 656 gallons if starting at a gross weight of 13234 and ending up with empty tanks
and a gross weight of 9300 pounds after flying the 2556 miles to Howland. This compares to the 680
gallons for the same conditions on figure VI.

Notice the maximum ranges calculated by the three different methods are quite similar.

A ten mph headwind component would have added about 200 statute air miles on the way to Howland
and a 25 mph headwind component would have added about 500 miles leaving more than 692 statute
air miles remaining even if the plane had left with only 1050 gallons.

So it doesn’t make sense that she ran out of fuel shortly after 2012 Z. Long’s explanation also doesn’t
make sense since there was no reason for AE to add power to speed up into the headwind since she
didn’t need to stay on a schedule and could just arrive later. The Lockheed report shows that to obtain
maximum range you should increase your airspeed only about 6 mph for a 20 mph headwind so if AE
had decided to increase her speed and power it would have been just a small amount and should not
have consumed all of her fuel. However, you only do this if you were flying at the optimum speed for
range, max L/D, and it appears that Earhart was flying at a higher speed when their weight got lighter.
If you are flying above the max L/D speed then, for a headwind, you should slow down so that your
speed is closer to the max L/D speed to achieve maximum range.

(If you look at graph II in Report 487 you will see this illustrated for 16,500 pounds. However, the curves
on this graph for the lighter weights are not located properly, they need to be further to the left because the
optimum speed is less at the lighter weights, so don't rely on them.)

To get the most accuracy out of the Breguet formula, which is based on the weight change from
takeoff to tanks dry, you should allow for the loss of weight from burning the oil, there was a reason for
carrying those 75 gallons of oil. Unlike modern flat engines, round engines consume a great deal of oil.
The specification for the Wasp is .32 oz per hour per horsepower which doesn't sound like much
but it does add up. Another way to state this specification is .02 pounds per hour per horsepower. Since
the SFC is .46 this means that for every 23 gallons of fuel that is burned the engines also burn 1 gallon
of oil. Burn 1,100 gallons of gas and you also burn 47.8 gallons of oil weighing 358.7 pounds. An easy way to do
this is to simply add the .02 pounds per hour to the SFC, totaling the burn off of all the liquids on board.
If you do this you get approximately 129 miles greater range estimate from the Breguet formula.

Something to keep in mind when using the Breguet formula is that it is optimistic in that it makes no allowance for taxi and take off and for climbing at higher power settings with a higher SFC and also assumes that the plane is flown at all times at the correct airspeed that results in L/D max which requires constantly slowing down as fuel is burned off. Flying either faster or SLOWER than the targeted airspeed will result in a shorter range.

But, she said she only had a half hour left...............

gl
« Last Edit: August 18, 2011, 12:02:01 AM by Gary LaPook »
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Gary LaPook

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• Posts: 1624
« Reply #7 on: August 15, 2011, 02:23:47 AM »

H.A.C
And according to Newtonian Aerodynamics, a Bee can'fly!  Tell that to them the next time you disturb a hive full of them.

Do it the simple way:
The Lockheed report gives the range of the 10E as between 4100 and 4500 miles on 1200 gallons of fuel.  Use the midpoint 4300 miles and divide by 1200 and get 3.5833 miles per gallon.
The Lae to Howland distance is 2556 miles, divided by 3.5833 eequals 713.3 gallons which when subtracted from 1100 gallons leaves386.7gallons, multiplied by 3.5833 m/g gives 1386 miles, enough to go to Gardner and back 3 times.
Which of course we gcould have gotten by subtracting the distance Lae to Howland, 2556 or so from 4300 to get 1744.
Conclusion:  They had more than enough feul to fly to Gardner, land, and run the starboard engine to charge the batteries to operate the radio to send out distress calls.

------------------------------------------------------------------------

Further fuel analysis.

gl

Two more attachments
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Gary LaPook

• T5
• Posts: 1624
« Reply #8 on: August 15, 2011, 09:41:37 AM »

H.A.C
And according to Newtonian Aerodynamics, a Bee can'fly!  Tell that to them the next time you disturb a hive full of them.

Do it the simple way:
The Lockheed report gives the range of the 10E as between 4100 and 4500 miles on 1200 gallons of fuel.  Use the midpoint 4300 miles and divide by 1200 and get 3.5833 miles per gallon.
The Lae to Howland distance is 2556 miles, divided by 3.5833 eequals 713.3 gallons which when subtracted from 1100 gallons leaves386.7gallons, multiplied by 3.5833 m/g gives 1386 miles, enough to go to Gardner and back 3 times.
Which of course we gcould have gotten by subtracting the distance Lae to Howland, 2556 or so from 4300 to get 1744.
Conclusion:  They had more than enough feul to fly to Gardner, land, and run the starboard engine to charge the batteries to operate the radio to send out distress calls.
-------------------------------------

You forgot to allow for the wind. The range values shown in report 487
are statute air miles, the distance that could be covered in still air.
If the plane was flying at  true airspeed of 150 mph into a 150 mph
headwind it would still be able to fly the same number of air miles but
the number of miles covered over the ground would be zero. It's easy
to allow for this affect, just multiply the headwind component times the
time in flight and subtract from the total air miles range values.

All the values for range in Lockheed report 487, and those computed with
the Breguet formula, are for still air and we know that Earhart had a
head wind on the way to Howland. One of the theories has her turning around
and returning to Rabaul and the wind would have become a tailwind. But the
same analysis holds for any "turn around" theory. Lets see how this works out.
Let's make up some numbers to make this easier to see. Let's say we are
going to fly, in a no wind situation, out 3,000 miles and then return
for a round trip of 6,000 miles in a plane that cruises at 150 mph. To
fly out 3,000 miles will take 20 hours (3,000 ÷ 150 = 20) and then it
will take an additional 20 hours to fly back for a total of 40 hours in
a no wind situation so we put on enough fuel to fly for 40 hours. Now
let's do it again, this time with a 50 mph headwind on the way out which
becomes a 50 mph tailwind on the way back. Since the winds are equal,
the same headwind and tailwind, the winds should cancel out so it should
still take 40 hours for the round trip. The average ground speed will
still be 150 mph ((200 + 100) ÷ 2 = 150) so this makes sense. Let's
confirm this. Going out, the ground speed will be 100 mph, the 150 mph
airspeed minus the 50 mph headwind. At 100 mph it will take 30 hours for
the outbound leg (3,000 ÷ 100 = 30). When the plane turns around the
ground speed jumps up to 200 mph, the 150 mph airspeed plus the 50 mph
tailwind. At 200 mph it will only take 15 hours to fly the 3,000 miles
back to our starting position, (3,000 ÷ 200 = 15.) Now adding the 15
hours for the return flight to the 30 hours for the outbound leg makes
the round trip take 45 hours. Wait a minute, that is more than the 40
hours it would have taken in still air, does that make any sense? Did we
commit a math error?

After double checking the math we see that this is correct, it will take
an additional 5 hours to fly the round trip with an equal 50 mph
headwind and tailwind than it did in still air. If we started with fuel
for 40 hours on board we would have run out 5 hours before reaching our
starting point. Since, on the return leg, we will have a ground speed of
200 mph, this means we will crash 1,000 miles short.

gl
« Last Edit: August 18, 2011, 12:05:23 AM by Gary LaPook »
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Harry Howe, Jr.

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• Posts: 576
• Nuclear Physicist(Ret) Pilot(Ret) Scuba(Ret)
« Reply #9 on: August 15, 2011, 12:33:13 PM »

No, I didn't forget the wind!
Looking at the lower number for range in the Lockheed Report (4100 miles on 1200 gallons) I realized that  the residual range when arriving at Howland (assuming still air) would have been about 1375 miles (4100 miles minus 2600 miles equals 1500 miles times 1100/1200 equals 1375 miles;  Even with a constant wind of 45 mph IN THEIR FACES for the entire 20 hour flight (45 times 20 equals 900 miles equivalent).  1375 miles minus 900 miles leaves 475 miles.  The distance Howland to Gardner is about 405 miles.

The wind was irrelevant to the conclusion, i.e.  AE/FN, diverting to land at Gardner, had enough fuel to fly there, land, fix  the radio, transmit distress calls, run the atarboard engine engine to charge the batteries as necessary.

The modern day equivalent to Occam's Razor (Look it up) is KISS (Keep It Simple Stupid).
No Worries Mates
LTM   Harry (TIGHAR #3244R)

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h.a.c. van asten

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• Posts: 322
« Reply #10 on: August 15, 2011, 12:38:13 PM »

With means and averages one should be very careful if , per your example , the tracks are equally 3,000 miles long . If tracks are different you never encounter a problem if you take the arithmetical mean of speeds etc . But if tracks are equal a treacherous trap comes up : for equal tracks the harmonical , not the arithmetical mean must be taken . If you make good 2 x 3,000 mls equal tracks @ 100 versus 200 mph the harmonic mean weighs with hours per mile :  2 / (1/100 hpm+ 1/200 hpm) = 2 / 0.015 = 133 1/3  mph for the mean speed , unequal to and lower than (100 + 200)/2 = 150 mph . That is the reason why you go in  undershot with fuel for 40 hours :  6,000 mls / 133 1/3 mph = 45 hours . Very treacherous this arithmetical gimmick , especially fo those computing for a combat range in war time . If the up/down speeds are equal (zero wind) , you have no problem for equal tracks and you can take the arithmetical mean of speeds . If the up/down tracks are unequal , you can also take the arithmetical mean of speeds .
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h.a.c. van asten

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• Posts: 322
« Reply #11 on: August 15, 2011, 12:52:37 PM »

Excuse sir, but that is too much solo : keep in mind that any acceleration against resistance , p.e. by 5% does not ask for 1.05 x , but for (1.05)^3 = 1,16 x the initial fuel flow . In the Earhart case , airspeed was increased to mean 155 mph to acquire 137 mean cross country speed , a difference of 13% , or factor 1.13 . The consequent fuel flow acceleration thence increased to (1.13)^3 = 1.44 x initial , or with 44 % . 714 galls x 1.44 = 1,028 gals , leaving 22 galls 87-octane and maximum 25 galls (but probably 16) 100-octane . The 22 galls (87-oc for the trip) covers Earhart´s GMT 1912 signal : 1/2 hour left .
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Gary LaPook

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« Reply #12 on: August 15, 2011, 01:02:51 PM »

No, I didn't forget the wind!
Looking at the lower number for range in the Lockheed Report (4100 miles on 1200 gallons) I realized that  the residual range when arriving at Howland (assuming still air) would have been about 1375 miles (4100 miles minus 2600 miles equals 1500 miles times 1100/1200 equals 1375 miles;  Even with a constant wind of 45 mph IN THEIR FACES for the entire 20 hour flight (45 times 20 equals 900 miles equivalent).  1375 miles minus 900 miles leaves 475 miles.  The distance Howland to Gardner is about 405 miles.

The wind was irrelevant to the conclusion, i.e.  AE/FN, diverting to land at Gardner, had enough fuel to fly there, land, fix  the radio, transmit distress calls, run the atarboard engine engine to charge the batteries as necessary.

The modern day equivalent to Occam's Razor (Look it up) is KISS (Keep It Simple Stupid).

-------------------------------------------------

But you didn't put that in your original post.

If you look at my post it also shows that they should have had enough gas to get to Gardner so it doesn't dispute your conclusion, it only illustrates for others that you must allow for the wind in this kind of analysis.

----------------------------------

But, she said she only had a half hour left...............

gl
« Last Edit: August 15, 2011, 01:07:44 PM by Gary LaPook »
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Harry Howe, Jr.

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• Nuclear Physicist(Ret) Pilot(Ret) Scuba(Ret)
« Reply #13 on: August 15, 2011, 01:07:54 PM »

I believe that the Chater Report states that all tanks were filled except one 81 gallon tank that was for 100 octane fuel and it was half full (40.5 gallons).  The reason that the 81 gallon, 100 octane tank, was not filled (topped off) was because Lae had no 100 octane fuel.
No Worries Mates
LTM   Harry (TIGHAR #3244R)

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