Signaling ships from 100 miles away
From BearWiki
Claim: The curvature of the Earth means that you're not going to be able to signal a ship from 100 miles away.
In the context of flagging down rescue by a ship from a raft, Bear claims that stranded people have been rescued, signalling from over 100 miles away[1]. Whether or not people have been rescued from over 100 miles away with a little reflector (can't find a reference to back it up) is up for grabs, but that would be an incredible amount of loss of reflected light due to the use of a non-collimated beam. However, rescue from a ship at 100 miles away is impossible due to the curvature of the Earth.
In Fig. 1, R is the radius of the Earth (around 3955 miles on average[2]), h1 is the height of the rafter, and h2 is the height of the observer on the boat. The longest possible line between them, d1+d2, intersects the Earth at its closest point to the center of the planet, radius R, forming two right triangles. Pythagorean theorum states that the squares of the two short sides of a right triangle sum up to the square of the hypotenuse, so R^2 + d1^2 = (R + h1)^2 and R^2 + d2^2 = (R + h2)^2. This expands to R^2 + d1^2 = R^2 + 2*R*h1 + h1^2 and R^2 + d2^2 = R^2 + 2*R*h2 + h2^2. This simplifies to d1^2 = 2*R*h1 + h1^2 and d2^2 = 2*R*h2 + h2^2, and then to d1 = sqrt(2*R*h1 + h1^2) and d2 = sqrt(2*R*h2 + h2^2). Now, assuming that the signal on the raft is done at a height of 6 feet, this works out to d1 ~= 3.0 miles, meaning that d2 is 97.0 miles. To solve for the required height on the boat, we have 97 = sqrt(7910*h2 + h2^2), which simplifies to 0 = h2^2 + 7910 h2 - 9409. Solving via the quadratic equation, we get a height of 1.2 miles. The tallest mast of the world's tallest ship is only 88.5 meters (290 feet) tall[3], and needless to say, people do not ride at the top of the mast.
Signalling an airplane handles the curvature of the Earth problem just fine, but you still have to deal with the loss-of-light problem. The sun is not a point light source. In our sky, it appears to take up about 0.5 degrees. That is, if you picture looking at the sun from the mirror's perspective, the light from it is reaching you in a cone that's one-half degree wide. The mirror is small enough compared to the sun that it effectively functions as a point reflector, so you approximate reflecting the cone of light from the sun (slightly truncated) at your target. It doesn't matter what angle your reflector is at; the sun will always form a cone of the same radius.
Your starting area of your beam is the surface area of your reflector (AR) times the cosine of half the angle between the sun and the target (theta). If you picture this cone split in half in 2D, you have a wedge with angle 0.25 degrees extending out toward infinity. Tangent is opposite over adjacent, so the radius of this wedge at whatever it hits can be described as tan(0.25 degrees) = r / d, where r is the radius and d is the distance. Thus, r = tan(0.25 degrees) * d, or r = 0.00436 d. The area at this distance is AB = pi * r^2 (plus the initial area, but that is proportionally insignficant), so AB = 5.972e-5 * d^2. So, at 100 miles, we have a beam that is 0.5972 square miles in cross-sectional area. To compare to the original amount of light, let's make some assumptions:
- AR = 12 square inches (2.989e-9 square miles) -- a very sizable knife.
- theta ~= 0 degrees (nearly aligned, to be kind)
You find that the light is approximately 200 million times weaker than you'd get by looking at an ideal reflection close-up. So, how bright will this look? On a perfect day in the tropics, you could concevably get one kilowatt per square meter[4], or 0.645 watts per square inch -- i.e., ~7.75W. In light terms, you could think of this as a flashlight with an 80 watt incandescent light bulb in the end -- pretty blinding if you look at it up close. Divide by 200 million for the distance, and you have the equivalent of a flashlight powered by a 0.4 microwatt (not milliwatt) bulb. Think you could see that?
This all assumes an optimal reflector. If you have a less-than-optimal reflector, everything is worse. It also assumes a perfectly transparent sky -- not a reasonable assumption.
Distance matters greatly. Cut the distance down to 10 miles, and that 100xes the strength -- 40 microwatt. This like having a flashlight with a bulb about 1/5th the strength of a malibu light (2W incandescent bulbs). Hard to see, but potentially visible if you're lucky and have incredibly keen eyes. For 100 miles, though, it seems you'd need a massive mirror to be visible; a knife just wouldn't cut it.
[edit] Analysis
Oppose: Re, curvature: Bear never explicitly claims that the rescue was from a ship. Doesn't address visibility, though.
- "Support": At the time he mentions this he was specifically talking about being seen by passing boats/ships and did not mention being seen by someone on land/plane etc, so it is logical to come to the conclusion that he was still referring to an incident of being seen by someone on a ship.
Support: So... would this mythical rescuee was signaling a space craft? Seeing as how atmospheric drag begins at ~75 miles, that would be the only other option...
