Congressional UAP/UFO Hearings Part 2; Napkin Calculation

We continue from Congressional UAP/UFO Hearings Part 1.

Reminder. This discussion requires that tic tacs are alien spacecraft, or from some parallel universe or hyperspace related to “here”, which may be false.

The tic tac descends from 80,000 feet  to near sea level in less than a second, or a few seconds, without breaking a sweat, which is attributed to advanced technology. What about the air? If it is a physical object, it pushes air.  A napkin calc follows, with these simplifications:

  • Tic tac is cylindrical, 15 meters length by 5 meters diameter, with domed ends.
  • Long axis is aligned with  the direction of travel.
  • Weight of atmosphere above 80,000 feet is negligible.
  • Atmosphere is 100% monatomic nitrogen, no oxygen.
  • Collision of nitrogen  with the tic tac is perfectly elastic; the tic tac is not heated by the collision. The atom gains energy; the energy of the tic tac is unchanged.
  • The nitrogen behaves according to simple kinetic theory, while actually it would become a plasma.  It has a temperature.
  • The result is an estimate of the energy used, “work”, to push the atmosphere out of the path of the tic tac in the descent from 80,000 feet to near sea level.
  • The calc is for 80,000 feet in 1 second, which is then scaled.

Spoiler. We should see something resembling an elongated nuclear explosion. If you’re not math inclined, skip to the end table.

Air pressure at sea level is 14.7 pounds per square inch. This is the weight of a square column of air 1 inch wide, mostly contained below 80,000 feet.

14.7 lbs/inch^2 = 22800 lbs/meter^2  = 207,000 kilograms over the area of the tic tac end cap, which is 19.53 meter^2. This is the total mass of the air displaced by the tic tac in its journey from 80,000 feet.

To approximate the reduction of momentum transfer of the domed ends compared to flat ends, 207,000 kg is reduced by 1/3.

The atoms of what was an air column acquire momentum of

p = 2.24 * 10^10 kilograms–meters/second

The kinetic energy of this mass is

E_total = p^2 / 2m = 1.9 * 10^15  Joules

The number of moles of what was the air column = 6.9 * 10^6 moles.

Multiplying by Avogadro’s number, the number of atoms is

N = 4.15 *  10^30

The average energy of an atom after collision with the tic tac is

E_one atom = 1.9 * 10^15 / N = 4.6 * 10^-16 Joules

Where K = Boltzmann’s constant, and T  is the Kelvin temperature, the average kinetic energy of a single atom is

E = 3/2 * KT

The temperature is

T = 4.6 *10^-16 / ( 1.38 * 10^-23 )  =  2.2 * 10^7 degrees Kelvin

= 30 million degrees Fahrenheit

TNT equivalent:

One ton of TNT is equivalent to  1.9 * 10^15 Joules of energy

The former air column has acquired energy equivalent to 460,000 tons of TNT.


16 miles / second,  Mach 80,     30,000,000 F,   460,000 tons TNT

8 miles / second,  Mach 40,        7.5 million   F,     125,000 tons TNT

4 miles / second,  Mach 20,        2 million F,            31000 tons TNT

The calc is useful for a craft traveling at the velocity of thermal atoms. At slower speeds, this approximation becomes inapplicable.

We studied the air because we know more about air than tic tacs. What have we learned?

To be continued shortly.