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