The Ladies and gentlemen! The Conspiromaniac continues holiday of the victory of the light of the knowledge against the darkness of political technology!
Excuse me, beside me little dirty. The Empty bottles, banks and the lacerated "corpses" of bodyguards of NASA wallow under legs.
But they guilty themselves - a weakling there is nothing to do on evening party of inflexible guys!
We ask to table! I have prepared you canard under name "ap16_salute". The tainted game is much in commissary under name NASA.
I have whetted its terrible knife and sliced the most luscious pieces for you. I have added its pepper and have roasted these pieces. Mmm? as it is tasty!
I offer from alcoholic drinks you vodka under name "Tears of astronauts are not flown to Moon".
I warn you it is very strong drink. I not advise for small babes and bodyguards of NASA to drink it otherwise they will have a strong emotional trauma.
With what do we begin? We begin with the tastiest certainly.
The First piece! He is identified "sand0.54". I have noted the red arrow showering sand with legs of the astronaut. The numeral in name of the piece means the time of frame. Slot is not seen else between sole of shoe of astronaut and sand.
The second piece of canard! He is identified "sand0.60". Here we see already that sand separated from of sole of shoe. The Red arrow points to upper edge of sand.
The Third piece! He is identified "sand0.73". Here sand nearly fell to the ground. Call your attention, soil is divided on the heavy sand (the red arrow) which I research and on the light dust (the green arrow) which hinging in midair. Such division can not be in vacuum! This occurs in atmosphere uniquely.
The Fourth piece! He is identified "sand0.80". The Heavy part of soil completely fell to the ground. But light dust continues to hang an air (call your attention, the word ?an air? without quote!).
We shall calculate time of the fall of sand: t=0.80 ? 0.54=0.26sec. Duration of one frame is 0.03 sec.
We shall do the correction for mistake 1 frame = 0.03 sec. Full time of the fall of sand (or NASA! Ha-ha-ha!) t=0.26 sec. ± 0.03 sec.
Now we shall calculate the height of the sand fall. This calculates harder, than calculate a time.
We need a standard of the size. I have taken the vertical size of astronaut?s knapsack (beside 0,8 meters) for standard of the size (it is specified by yellow arrow).
The size of knapsack on video frame is 39 pixels.
Consequently scale is 0.8/39=0,02m/pixel.
Let's calculate the height of the fall of sand on the first frame "sand0.54".
178 ? 162 = 16 pixels, consequently, h=16*0.02=0.32m.
Let's calculate the speedup of the free fall of sand: g=2h/ (t*t); g=2*0.32/ (0.26*0.26)=9.47m/s2 (Ur-ra!!!)
Let's consider with inaccuracy of the measurement of time ± 0.03 sec.
With inaccuracy (-0.03sec.) g=2*0.32/ (0.23*0.23) =12.1 m/s2
With inaccuracy (+0.03sec.) g=2*0.32/ (0.29*0.29) =7.6 m/s2
Moon speedup of the free fall 1.62 m/s2 does not fall into this interval from 7.6 before 12.1!