Bah. So a good friend sent us a link to a science-y blog with some big equations that apparently prove the Mystery Missile off California's coast this week is in fact...wait for it...a plane.Since we're usually dumbfounded by fancy equations and by science folks in general we'll call this one done.
Too bad, a North Korea nuke boat with an errant missile off California's coast would have been a lot more fun.
Back to sharks after this:
This diagram is not to scale, but the math is the same regardless. The solid curved line is the surface of the earth. The dot at the top is San Clemente. The little triangle is Catalina. “d” is the distance to Catalina (d=35 miles). “c” the amount of Catalina that is visible above the horizon (c=0.05 miles, really a bit more, but let’s be conservative). “a” is the altitude of the plane, (a = 6 miles). “r” is the radius of the earth (r=3963 miles).
The green wavy line is the contrail. Notice it’s at a fixed height above the surface of the earth, and is going directly towards the OC.
The point labeled (0,0) is the center of the earth. (0,0) means X=0, Y=0, where X is horizontal and Y is vertical. What we want to know is how far away the plane is, the value x. We do this with cartesian geometry, noting that the lowest visible point of the trail is at the intersection of the dotted line, which is a circle of radius (r+a), hence the equation x^2 + y^2 = (r+a)^2 and the line labeled “sight line”, which is has the equation y=x*c/d. Combining these equations to solve for x yields a quadratic equation, which we can solve with Wolfram Alpha:
intersection of (y=r+x*c/d) and (x^2+y^2 = (r+a)^2)
and with the real numbers:
intersection of (y=r+x*c/d) and (x^2+y^2 = (r+a)^2) where a=6 and d=35 and c=0.05 and r=3963
Thanks to Ethan Stock for the hat tip!
