Gary, please allow me another question related to take off distances. If Lae to Howland used almost 1890 feet of runway due to the weight of the plane and a full fuel load, then does that mean it would be the same approx 1890 feet taking off from Howland to Lae. As she would have done on the first world attempt? If yes then would she have used the longer runway? Remembering that the 300 foot extension wasn't there for the first attempt. Purely from the standpoint of interest. It has no bearing on the hypothesis. Thank you in advance.

I don't know where you got the 1890 foot value from. My prior calculation for takeoff at Lae was 2278 feet without flaps and 1914 feet with 30° flaps. This was at sea level, 85° F, no wind, 1100 gallons of fuel on board making the weight 15,000 pounds. The actual no flap takeoff distance measured by Chatter was 2,550 feet, only 272 feet more than the calculated distance. If you don't believe Chatter then you at least know that the plane could take off in less than 3,000 feet under those conditions.

See my prior post here. Since the weather conditions are virtually the same if taking off from Howland to fly to Lae we know that the plane could certainly get off in 3,000 feet and most likely in 2,550 feet or a little bit less without flaps. The north-south runway at Howland was much longer than 3,000 feet and there is no reason to believe that the plane could not take off with a 15 mph crosswind on that runway. If taking off into a 15 mph east wind then the 2,550 feet becomes only 1,867 without flaps and the 1,914 foot flaps down takeoff would only take 1,349 feet so either way the east runway would have been long enough.

However, flying to Lae from Howland with the existing, fairly normal, wind conditions on that route the flight would take significantly less time so the plane could have taken off with less fuel. It is 2,556 SM miles and the 25 mph headwind on the leg from Lae to Howland would be a 25 mph tailwind going the other way making the ground speed 50 mph faster and taking about 5 hours less. This means that that flight could be made with about 250 gallons less fuel which weighs 1,500 pounds so the takeoff could have been made at a significantly lower weight of 13,500 pounds which is approximately the same as I calculated for the Howland to Hawaii flight.

I wrote before:

"Except we know that they had planned to carry only 825 gallons of fuel for the leg from Howland to Hawaii which is only 1892 SM, 664 SM shorter than the Lae to Howland leg. This is 275 gallons less than they carried on takeoff from Lae so the weight of the plane would have 1650 pounds less, only about 13,350 pounds. Doing the same takeoff calculations I have shown before we know that the takeoff distance varies with the weight ratio squared. Report 487 gives 2100 feet for a takeoff at 16,500 pounds, 13,350 pounds is only 0.809 of the maximum weight, which squared equals 0.654 times 2100 feet predicts the takeoff distance at Howland of 1,375 with 30° of flaps. For a no flap takeoff the distance would be the same 0.654 times 2,600 feet for a Howland takeoff distance of 1,700 feet."

Then adjusting this for a takeoff into the wind, I wrote before:

"From my prior post, the takeoff distance with no wind at Howland is 1,375 feet with 30° of flaps. For a no flap takeoff the distance would be 1,700 feet. A 20 mph headwind would shorten the 30° flap distances by 41.5% and a 10 mph headwind would shorten these distances by 22%. So with 30° of flaps and a 20 mph wind the distance would be 804 feet, with a 10 mph wind it would be 1070 feet. Without flaps, the reductions would be 38% and 20% so the distances would be 1050 and 1356 respectively.

BTW, I don't make this stuff up, see attached graph from Aerodynamics For Naval Aviators, the official Navy manual.

The calculation is easy. The takeoff speed of the plane with the lighter fuel load of 825 gallons for the shorter flight from Howland to Hawaii at the gross weight of 13,350 with no flaps is 93.5 mph and for a 30° of flaps takeoff the takeoff speed is 85 mph. The formula for computing the takeoff distance allowing for wind is:

Takeoff distance (wind) = Takeoff distance (no wind) (1 - wind speed/takeoff speed)^2

So for the 30° flap takeoff with a headwind component of 20 mph the calculation looks like this:

TD(w20) = 1375 (1- 20/85)^2

TD(w20) = 1375 (1 - 0.235)^2

TD(w20) = 1375 (0.585)

TD(w20) = 804 feet

You can work out the other examples yourself."

You can do the calculation for a 15 mph wind.

gl