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Author Topic: LaPook Hypothesis: Box Search around 157-337  (Read 122846 times)

John Hart

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Re: LaPook Hypothesis: Box Search around 157-337
« Reply #105 on: July 08, 2012, 07:31:37 PM »

Wow, it's great to see my bar napkin numbers are close to Gary's on his attached charts.  Left out Rawaki by the way, you only show the span to Kanton.  The whole cluster of islands is hardly a "fly speck".

Doesn't matter how long someone says, or we think, the airplane will float, only how long FN thinks it will float.  Remember he has been through a lot of seaplane landings, probably a few rough ones, and those were in lagoons.  He knows open water landings are avoided by them.  I would expect he would do anything to avoid having to ditch open ocean.  I believe I read somewhere no liferaft on board.

Again, you are arguing navigation techniques not pilot judgement.  What's your plan when you get to the end of your gas after all these pretty rectangles?  Why if you kept flying in and out of range of Itasca on this search pattern did you not communicate with them again?  Who was looking, who was flying, who was navigating, and from where?  How were you communicating with each other if opposite ends of the airplane?

But once again I will say, you may be right.  But so too may others.  I'll give you yours, why can't you give others theirs?  I can shoot holes in yours, you can shoot holes in mine but we don't need to impinge each other's qualifications to have a theory.  I give you yours.  COAs are exactly that, choices.  FN and AE had choices.  We don't know which they made but we can speculate.

The world wonders...

JB
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JNev

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Re: LaPook Hypothesis: Box Search around 157-337
« Reply #106 on: July 09, 2012, 08:14:06 AM »

...For me, the more cogent question is "Why would AE/FN abandonestablished "best practices" and head off looking for a flyspeck in the broad Pacific?"  Such a thing COULD have happened but if it did we need a "why".

William
#3425

He wouldn't have had to 'abandon established best practices' to 'head off looking for a flyspeck' -

Go to Gary's site about 'fredienoonan' and read thoroughly among the navigational texts he's posted there: not only is the search pattern discussed, so is alternate landfall when a search fails.  It takes a bit of reading but the point is discussed among those texts (I recall reading it there but am too lazy to find it at moment).

As to 'flyspecks' - Howland was more nearly a flyspeck than Gardner: big difference in appearance.

"Why"?  A hope for dry feet, perhaps; dumb luck, may well have been.  We have to find that they actually got there before we can hope to draw near a real "why", though, and may never know.

Would FN have known that Gardner was more than a 'fly speck'?  Apparently not from his charts as then available.  I think that's been one of Gary's points and it is a fair one. 

That leads to my own thought that IF Gardner was the place of landing, how they got there will likely always be a mystery.  It could have been dumb luck for a crew that was severely lost and in the throws of flying a box search for Howland after all... or, it could have been a lucky stumbling into the proverbial fly speck.  I don't think it can ever be proven.

Gary's work remains fascinating - the guy is a font of navigational stuff!

LTM -
- Jeff Neville

Former Member 3074R
 
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Jeff Victor Hayden

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Re: LaPook Hypothesis: Box Search around 157-337
« Reply #107 on: July 09, 2012, 08:34:43 AM »

Fred was a smart guy, he knew the risks. I have no doubt a search pattern was implemented but, to search until zero gas knowing the inevitable outcome of not locating Howland/baker doesn't sound smart. Would Fred be smarter than that with all his experience of navigation and, his work with Pan Am?

Smart route out of an unkown situation, play the percentages.

Part one, a limited search pattern/time to find two dots of land, Howland/Baker
Part two, head towards eight dots of land, The phoenix islands.

Two dots of land versus eight, playing the percentages?
I know the Phoenix Islands, 8, were not a 'catchers mitt' but then again, in comparison to Howland/Baker, 2, ????
How much time and fuel would you devote to looking for 2 when you could be looking for 8 ?


This must be the place
 
« Last Edit: July 09, 2012, 01:30:37 PM by Jeff Victor Hayden »
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Martin X. Moleski, SJ

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Re: LaPook Hypothesis: Box Search around 157-337
« Reply #108 on: July 12, 2012, 12:07:03 PM »

I've just had an epiphany in another thread.

I'm hauling this piece over here where it belongs, more or less.

Suppose AE and FN stayed in the air, doing a LaPook search from the south while the Itasca headed northwest.  The absence of the ship and the smoke (however useful it might have been) would reduce the visibility of Howland.  It might not be hard to get the endpoint of the LaPook search within five to seven miles of Howland.  As the Itasca steams away (full speed ahead?) to the northwest, AE and FN work their way closer and closer to Howland from the south.
LTM,

           Marty
           TIGHAR #2359A
 
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Jeff Victor Hayden

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Re: LaPook Hypothesis: Box Search around 157-337
« Reply #109 on: July 12, 2012, 12:21:49 PM »

Quote
AE and FN work their way closer and closer to Howland from the south.

Which is logical and, at what stage of this search pattern do they decide that it's a waste fuel trying to find the two dots (still not in sight, even more so now as Itasca has headed off in the opposite direction) and head towards the eight dots?
And of course if Itasca did guess correctly and NW was the place and AE and FN have got their search pattern up and running they would be closing in on each other on the 'correct' side of the search pattern.
This must be the place
 
« Last Edit: July 12, 2012, 12:58:00 PM by Jeff Victor Hayden »
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Gary LaPook

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Re: LaPook Hypothesis: Box Search around 157-337
« Reply #110 on: July 14, 2012, 01:00:18 AM »

I think we have all been making the mistake of believing that the visibility reported by Itasca was also the distance that Howland could be spotted. These distances are not the same. To use the CAP terminology, Itasca reported "search visibility" while the distance we can expect to spot the object we are looking for is defined by the CAP as "scanning range." People have also confused the "geographic range," the distance to the horizon based on your altitude. (1.144 times the square root of your altitude in feet, 1000^.5 = 31.6 X 1.144 = 36.2 NM) with scanning range. Here is how we figure out the geographic range, boys and girls. The first step in the process is to calculate the square root of 1,000 so take out your slide rules and line up the hair line on the  cursor with the "1" in the center of the "A" scale. Now look at the "D" scale and you see that the hair line is lined up with 316. To place the decimal point we do a rough approximation in our heads. Ten times ten equals one hundred so we know the square root of 1,000 must be greater than ten. One hundred times one hundred is 10,000 so we know that the square root of 1,000 must be between  10 and 100 so we can now place the decimal point in the proper location making the square root of 1,000 equal  31.6. Now, without allowing the cursor to move, move the slide to the right until the "1" on the "C" scale at the left end of the slide is lined up with the hair line, then move the hairline to line it up with 1.144 on the "C" scale and then read out the geographic range under the hair line on the "D" scale, which is "36," making the geographic range 36 NM. The constant, 1.144, is derived from the size of the earth and the definition of the nautical mile. For those of you without your slide rules today, take out your table of logarithms and extract the mantissa of the "1" in "1,000." O.K. that was a joke because everyone knows that the mantissa of "1" is zero. Now place the characteristic of "3" in front of the "0" and we have the complete logarithm of 1000 which is 3.00000. To find he square root we divide this logarithm by "2" making the logarithm of the square root of 1,000 as "1.50000." We then scan the logarithm table for the mantissa of "50000" and take out the antilog of "3162." Since we know the characteristic is "1" we can place the decimal point and so determine the square root of 1,000 is 31.62 NM. But the beauty of logarithms is that we can skip this step since we are not really interested in the square root, but the square root multiplied by 1.144. So, we simply extract the logarithm of 1.144, which is  0.058426, and add this to the 1.50000 making it 1.558426. Now, looking up "558426" in the mantissa table we extract "3618" and placing the decimal point based on the characteristic of "1" we now have the geographic range of 36.18 NM.

So what is the scanning range for Howland island? Human visual acuity has several facets but the most common aspect is measured by the standard Snellen test hanging on your doctor's wall:
        E
     F P

     T O Z

     L P E D

(these are the most common lines, I memorized them a long time ago.) Standard visual acuity is listed as 20/20 and the line that results in this score has the letters of a size that the thickness of the characters and the spaces between the elements in each letter subtend an angle of one minute of arc, 1/60th of a degree and this is in the doctors office where the air is clear and with a standard, high level of illumination. Some people can see better than this and some objects that are much smaller, like stars (due to the high contrast) can also be seen but the one minute of arc acuity is a good value for us to use in thinking about the scanning range for Howland. The sine (or tangent, it doesn't matter) of one minute of arc is 0.00029 the inverse of which is 3438 which means that the ratio is 1 to 3438.  A common approximation of this value is 1 to 3600 which is one inch high at a distance of 100 yards (3600 inches) and this is used universally by rifle shooters and is close enough for our purposes. This means that a person with 20/20 vision should be able to see something that is one inch high at 100 yards and not be able to see anything that is smaller at that distance. As an experiment I put some pieces of orange tape that were two inches wide on the back of a road sign (see attached photo) and then drove away about a half mile and then drove towards the sign slowly and then stopped when I could first detect the orange tape. I measured the distance with my laser rangefinder and it was 212 yards which is slightly more than the expected 200 yards but it was a highly contrasting color and I knew where to look.

An object that is one foot tall should be visible at 3,600 feet which is about 6/10 ths of a nautical mile. The highest point on Howland island is only 18 feet high so should not be visible more than 10.8 NM away (10.2 NM if you do the actual trig) no matter what the geographic range or the reported visibility (unless those restrict it even more.)

But that is not the end of the story since this just describes the situation of a person looking for Howland from shipboard and Earhart was looking from 1,000 feet above sea level. Think of Howland as a playing card. I have just described the situation of one looking at the card edge on. Obviously you can see the card farther away if you are looking at the front of the card instead of the edge. Howland is about one-half NM wide so how close would we have to be for that 1/2 NM wide island to subtend one minute of arc from a point a thousand feet higher than the island? Think of the playing card laying flat on a table with the short dimension towards you. Since you are taller than the table you will be looking down on the card but from all the way down the hall the card would subtend a small angle and would  look quite narrow to you if you could see it at all. As you move closer to the table the card will look bigger and bigger and when you are looking straight down on the card is will look its biggest. When you get close enough so that it is seen to subtend one minute of arc you should be able to see it.

Doing this computation for Earhart we compute the angle the plane would appear above horizontal from the near shore and from the far shore of the island and take the difference, this would be the angle the island would subtend from the plane. At 1000 feet high and 20 NM away (the near shore), the plane would be 28.3' above the horizon while from 20.5 NM away (the far shore) it will be 27.6' high a difference of only 0.7' so not subtending an angle large enough to be seen. From 18 NM the angle is 31.4' and from 18.5 NM it is 30.6' a difference of only 0.8', still not big enough to be seen. From 16.5 NM the angle is 34.3' and from 17 NM it is 33.3' a difference of 1.0' so should be able to be seen. (See attached diagram) So this should be the maximum distance that the island could be seen, not the 20 NM everybody has been using for their computations. Of course this assumes clear air and good contrast so it is more likely that the island could not actually be seen until closer than this theoretical 16.5 NM

What about spotting the Itasca?
Fortunately we have good scientific information about searching for ships in the National Search And Rescue Manual which uses the concept of "sweep width" instead of scanning range. I have attached table 4-5 which shows the sweep width when searching for a ship between 150 and 300 feet long from an altitude of 1,000 feet the sweep width is 24.7 NM with 20 NM visibility and 34.9 NM if the visibility is 30 NM. This is the new scientific search method. So what does this mean? With 20 NM visibility, one pass of the search plane covers a swath 24.7 NM wide, 12.35 on both sides of the plane and there is an 80% probability that the ship will be spotted if it was within that swath. So one way to interpret this is that Earhart had an 80% probability of spotting the Itasca if they flew within 12.35 NM of the ship. So if you wanted to achieve an 80% probability of spotting this ship then you should plan your search pattern so that the legs are spaced one sweep width apart and when this is done the "coverage factor" is one. If we want a higher probability of success we just have to increase the coverage factor either by searching the same track multiple times or by spacing the search pattern legs closer together. I have attached table 5-19 which shows the cumulative probability of detection based on coverage factor. We can see that with a coverage factor of one (the track spacing equals the sweep width) on pass results in an 80% POD while two passes on the same track increases it to 95% and three passes to 99%. If we space the search pattern at half the sweep width then the coverage factor becomes 2.0 and the POD increases to 98% for one pass. This means that we have a 98% probability of spotting that ship if we pass within 6.175 NM, half the distance of before. We can also use this table to see what happens if we pass farther away from the ship. If we pass twice as far away, 24.7 NM, then the coverage factor is 0.5 and the POD for one pass drops to 47% and for two passes it is only 72%. What would be the probability of spotting Itasca 20 NM away with 20 NM visibility? Divide 12.35 by 20 and you find the coverage factor is 0.62 so the POD for one pass is 55% and 80% for two passes. Something we can learn from this is that when searching for a ship this size, if you equate the visibility with "scanning range" then you have only about a 50-50 shot at spotting the ship. So if they were flying a search pattern with the legs spaced 40 NM appart, twice the visibility reported by Itasca, then they would only have had a 55% probability of spotting itasca and a low probability of spotting Howland.


So the problem might be that Earhart and Noonan overestimated the range that they could spot Howland.

gl
« Last Edit: July 16, 2012, 12:19:31 AM by Gary LaPook »
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John Ousterhout

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Re: LaPook Hypothesis: Box Search around 157-337
« Reply #111 on: July 14, 2012, 06:41:59 AM »

Gary,
Do you know of any guidelines for searchers when the sea surface is dotted with cloud shadows, of approximately the size of Howland in AE's case?  Similarly, are there guidelines for searching for a downed aircraft in a sea with whitecaps that also approximate a floating Electra in size, that is also dotted with cloud shadow?
Cheers,
JohnO
 
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Bruce Thomas

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Re: LaPook Hypothesis: Box Search around 157-337
« Reply #112 on: July 14, 2012, 07:25:09 AM »

People have also confused the "geographic range," the distance to the horizon based on your altitude (1.144 times the square root of your altitude in feet, 1000^.5 = 31.6 X 1.144 = 36.2 NM) with scanning range.

Gary! Groan!  You're too good with math for me (who teaches undergraduate math in a local university) to let that one go without comment.  When my students write gibberish like "1000^.5 = 31.6 X 1.144 = 36.2" they get the dreaded red pen markup.  Because "things equal to the same thing are equal to each other" ... so you're in effect saying that the square root of 1000 is 36.2, and I know you know better. 

One way to write it correctly (and to continue to infer that your readers know that you're giving an example) with an altitude of 1,000 feet:  (1000^.5 = 31.6) X 1.144 = 36.2

Your friendly math pedant.   :)
LTM,

Bruce
TIGHAR #3123R
 
« Last Edit: July 14, 2012, 08:18:36 AM by Bruce Thomas »
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Jeff Victor Hayden

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Re: LaPook Hypothesis: Box Search around 157-337
« Reply #113 on: July 14, 2012, 07:36:37 AM »

We may have inadvertently stumbled upon the reason why no one could see each other. During the Itasca search and, the following larger scale search everyone was studying tables and working out mathematical formulae instead of looking over the side of the plane or ship
 ;)

This must be the place
 
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C.W. Herndon

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Re: LaPook Hypothesis: Box Search around 157-337
« Reply #114 on: July 14, 2012, 07:43:05 AM »

People have also confused the "geographic range," the distance to the horizon based on your altitude (1.144 times the square root of your altitude in feet, 1000^.5 = 31.6 X 1.144 = 36.2 NM) with scanning range.

Gary! Groan!  You're too good with math for me (who teaches undergraduate math in a local university) to let that one go without comment.  When my students write gibberish like "1000^.5 = 31.6 X 1.144 = 36.2" they get the dreaded red pen markup.  Because "things equal to the same thing are equal to each other" ... so you're in effect saying that the square root of 1000 is 36.2, and I know you know better. 

One way to write it correctly (and to continue to infer that your readers know that you're giving of example) with an altitude of 1,000 feet:  (1000^.5 = 31.6) X 1.144 = 36.2

Your friendly math pedant.   :)

Way to go Bruce. :)

Sometimes Gary gets so technical in his explanations that I just skim through them. I sure missed that rather obvious error. :o
Woody (former 3316R)
"the watcher"
 
« Last Edit: July 14, 2012, 07:45:20 AM by C.W. Herndon »
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C.W. Herndon

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Re: LaPook Hypothesis: Box Search around 157-337
« Reply #115 on: July 14, 2012, 07:53:18 AM »

We may have inadvertently stumbled upon the reason why no one could see each other. During the Itasca search and, the following larger scale search everyone was studying tables and working out mathematical formulae instead of looking over the side of the plane or ship
 ;)

Jeff, you may be on to something here. :o
Woody (former 3316R)
"the watcher"
 
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Martin X. Moleski, SJ

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Re: LaPook Hypothesis: Box Search around 157-337
« Reply #116 on: July 14, 2012, 09:23:44 AM »

We may have inadvertently stumbled upon the reason why no one could see each other. During the Itasca search and, the following larger scale search everyone was studying tables and working out mathematical formulae instead of looking over the side of the plane or ship
 ;)

It seems to me that Fred faced a tradeoff during the search.

He could have his attention directed inside the cockpit or outside, but not both.

Time spent on navigational calculations would be time stolen from doing the visual search, and vice-versa.

I'm sure Gary can calculate how long Fred could spend looking down at a chart before running the risk of not scanning his side of the search path effectively.  That time could be compared to the time needed to calculate something--say, making an estimate of distance and course covered to update the dead-reckoning or calculating wind-drift so as to give AE the right heading to fly to cover the desired search path.

That seems to me to be a possible contributing factor to them winning the "Close, but no cigar" award for their flight.
LTM,

           Marty
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Gary LaPook

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Re: LaPook Hypothesis: Box Search around 157-337
« Reply #117 on: July 14, 2012, 11:12:41 AM »

People have also confused the "geographic range," the distance to the horizon based on your altitude (1.144 times the square root of your altitude in feet, 1000^.5 = 31.6 X 1.144 = 36.2 NM) with scanning range.

Gary! Groan!  You're too good with math for me (who teaches undergraduate math in a local university) to let that one go without comment.  When my students write gibberish like "1000^.5 = 31.6 X 1.144 = 36.2" they get the dreaded red pen markup.  Because "things equal to the same thing are equal to each other" ... so you're in effect saying that the square root of 1000 is 36.2, and I know you know better. 

One way to write it correctly (and to continue to infer that your readers know that you're giving an example) with an altitude of 1,000 feet:  (1000^.5 = 31.6) X 1.144 = 36.2

Your friendly math pedant.   :)
O.K., I went back and fixed it.

gl
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Bruce Thomas

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Re: LaPook Hypothesis: Box Search around 157-337
« Reply #118 on: July 14, 2012, 12:02:21 PM »

O.K., I went back and fixed it.

I'm so proud!  Two bonus points!  :D
LTM,

Bruce
TIGHAR #3123R
 
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Jeff Victor Hayden

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Re: LaPook Hypothesis: Box Search around 157-337
« Reply #119 on: July 14, 2012, 12:15:12 PM »

O.K., I went back and fixed it.

I'm so proud!  Two bonus points!  :D

Bruce, I never thought I would see the good old T183 again. It was the calculator we had to purchase for the mathematics modules on our degree course as it was extremely programable and you could compile and run simple software programs that you had created. The statistical functions were outstanding, ideal for this thread.
I wonder if it's still in a cardboard box in the garage?
This must be the place
 
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