UFO talk. I saw some talk about the “tic tac” a few days ago. As reported, it’s… implausible that it was a physical object.
Math time! I’m taking the numbers from the wiki page of the incident, as reported by Commander Fravor.
the object had now dropped from 28,000 feet to near sea level in less than a second
What can that tell us about how this thing moves?
28,000ft = 8,534m
8,534m/1 second / 343m/s = Mach 24.88
Right off the bat, we’re looking at something that can do ~Mach 25 in atmosphere. But that’s dramatically underselling it - the Space Shuttle would start reentry around that speed. We’re not taking acceleration into account.
Sighting similar objects on radar a few days prior, Officer Day on the Princeton “was startled by their slow speed of 100 knots”. So let’s assume the tic tac didn’t have any vertical velocity to start, and didn’t pancake into the ocean at Mach 25. We’ll see that 100 knots rounds down to nothing.
So our UFO needs to start with 0 velocity, accelerate downward for half a second, instantaneously reverse, and accelerate the same amount in the opposite direction for another half a second, dropping 8,534m in the process.
We can select the appropriate kinematic equation to get acceleration, given we know displacement, initial velocity, and time.
What do we get in the first half second?
Δx = v_initial * t + 1/2 * a * t^2
4,267m = 0 * .5 + 1/2 * a * .5^2
Solving for acceleration gives
a = 4,267 * 2 * 4 = 34,136m/s^2
Dividing that by 9.81 gives 3,479 gees. After half a second, it’s traveling 17km/s, Mach 99.
Given what we know about physics, doesn’t seem likely that anything could perform that kind of maneuver, especially in atmosphere.