Gary

Thanks for the critique, two things:

1. Why would they fly a NNW offset? There's nothing up there (unless you include the Japanese mandated islands.

2. Let the leg from Lae to a NNW offset point be A, a leg of a triangle, and B the sffset distance be another leg of the triangle, and C the direct distance Lae to Howland be the hypotenuse. How can the distance A plus B be less than C?

The speed difference between 159 and 162 is less than 2%. A time increase of one-half hour in a trip of 22 hours is about 2.3%. The Electra had a cruising speed capability of 190 mph and a never exceed speed of 202mph so they had plenty of margin to adjust their speed as they progressed along their course, taking sightings and comparing their progress with what they wanted in order to arrive at a LOP around sunrise , which I believe was 0547 Howland.

Harry, back in September I invited you do draw a diagram so that you would be able see that an intercept to the north-northwest was shorter than an intercept to the south-southeast but you apparently have not done so. See:

See:

https://tighar.org/smf/index.php/topic,452.msg5640.html#msg5640and:

https://tighar.org/smf/index.php/topic,452.msg5646.html#msg5646So to make it easy for you to see this, I am attaching two diagrams that I have drawn for you. Diagram "A" shows the course from Lae to Howland, 2556 SM long, and the courses to intercept points 202 SM out, both to the north-northwest and to the south-southeast. I exaggerated the angles between the course from Lae to Howland and the LOP for clarity. You should be able to see the answer to your question that the distance to the offset point to the north-northwest

**is shorter** than the direct course to Howland and the distance to the other offset point is longer.

Diagram "B" is drawn to scale and you should still be able to see the difference in the two distances.

The reason to fly an offset to the north-northwest is to be able to find Howland, they have no interest in going anywhere else.

No they can't just go a lot faster just to meet your idea that they should intercept the LOP at sunrise, 202 SM out to the right. To go faster requires much more power, it increases as the cube of the airspeed, which then requires full rich mixture (you should know this) so the BSFC goes way up and the specific range goes way down. This means that you get much fewer miles per gallon so you can run out of fuel before you get to your destination.

There is no reason to try to keep to your schedule because you cannot take an accurate sextant reading from an airplane at the time of sunrise because the sun's altitude is actually below zero and the Pioneer octant does not have a scale below zero. Even if it did, Noonan still would not be able to take an accurate sextant observation until the sun was six degrees above horizontal because the tables Noonan was using did not provide correction factors for lower altitudes. It took about one-half hour for the sun to climb above six degrees. The optimum time to reach the LOP was about an hour after sunrise as this would allow about a half-hour period when the sun was high enough for observations so that Noonan could work out his ground speed and the ETA at the LOP.

See:

https://sites.google.com/site/fredienoonan/discussions/the-myth-of-the-sunrise-lopYou can also work it out using the law of cosines.

Intercepting to the NNW:

C^2 = A^2 + B^2 -(2 AB cos c)

C^ = 2556^2 + 202 ^2 - ( 2 x 2556 x 202 x cos

**79°**)

C^ = 2556^2 + 202 ^2 - ( 2 x 2556 x 202 x

**0.1908**)

C^ = 6533136 + 40804

**-** ( 197033.948)

C^2 = 6376906.052

C =

**2525.254**Intercepting to the SSE:

C^2 = A^2 + B^2 -(2 AB cos c)

C^ = 2556^2 + 202 ^2 - ( 2 x 2556 x 202 x cos

**101°**)

C^ = 2556^2 + 202 ^2 - ( 2 x 2556 x 202 x

**-0.1908**)

C^ = 6533136 + 40804

**+** ( 197033.948)

C^2 = 6770973.948

C =

**2602.11**The cosine of 101 degrees is a

**negative** .1908 so the second term ends up being a plus which is why this intercept is longer.

gl