This has been rumoured for some time, but NASA has finally announced officially that later this week, in the run-up to the 40th anniversary of the Apollo 11 landing on the Moon, it will be holding a big media event – and broadcasting it live, on NASA TV – to show some “new” footage of the Apollo 11 moonwalk.

Of course, it’s not “new” footage at all – it’s not as if they’ve found film taken by a previously unknown camera carried to the Moon onboard Eagle! No, reading the press release, what is being shown on Thursday is a new version of the footage we already know about, an enhanced and better version of the footage we’ve already seen. It’s probably been cleaned up and clarified and tweaked to within an inch of its life using modern processing techniques.

If that’s the case, what will it show? Will we see Armstrong’s face in his visor more clearly? Will we see that historic First Step in more detail? I guess we’ll just have to wait and see! In the meantime, here’s the text of the press release…

NASA Holds Briefing to Release Restored Apollo 11 Moonwalk Video

WASHINGTON — NASA will hold a media briefing at 11 a.m. EDT on Thursday, July 16, at the Newseum in Washington to release greatly improved video imagery from the July 1969 live broadcast of the Apollo 11 moonwalk.
http://www.nasa.gov/ntv

The release will feature 15 key moments from Neil Armstrong’s and Buzz Aldrin’s historic moonwalk using what is believed to be the best available broadcast-format copies of the lunar excursion, some of which had been locked away for nearly 40 years. The initial video released Thursday is part of a comprehensive Apollo 11 moonwalk restoration project expected to be completed by the fall.

The Newseum is located at 555 Pennsylvania Ave. N.W. The news conference will be broadcast live on NASA Television and streamed on the agency’s Internet homepage.

Participants in the briefing will be:

— Richard Nafzger, team lead and Goddard engineer

— Stan Lebar, former Westinghouse Electric program manager

— Mike Inchalik, president of Lowry Digital, Burbank, Calif.

For NASA TV downlink information, schedule information and streaming video, visit:

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Gordan, on July 13, 2009 at 8:12 pm said:I think they’re talking about that grainy, black/white TV footage and not the 16mm film footage your inset shows. The latter one is pretty much as good as it gets, but the TV footage usually circulated around is horrendous, n-th generation copy. Allegedly, there were tapes made directly at the DSN station in Australia that are more crisp and don’t suffer from 37 conversions on their way to your screen, but these are probably not those but something closer to the source.

We shall see…

Siegfried Marquardt, on January 17, 2014 at 8:45 pm said:Refutation of Apollo 11 on the basis of the fuel balance for the start and flight trajectory

1 Representation of NASA for launch of Apollo 11 in the cosmos

According to the website de.wikipedia / org July 2013, the version of NASA is to the launch of Apollo 11 in the cosmos as follows (quoted from Wikipedia ) : “Apollo 11 launched on 16 July 1969 at 13:32:00 UTC at the top of the 2940 -ton Saturn V from Cape Canaveral , Florida, and reached Earth orbit as planned twelve minutes later . After one and a half orbits , the third stage of the rocket was ignited again. It burned for about six minutes and took the Apollo spacecraft on the moon course. A short time later, the Command / Service Module (CSM ) was coupled to the lander . ” (Quoted from Wikipedia for Apollo 11, 2013). How could please work because this beautiful astrophysical and so can happen ? The Apollo spacecraft was launched lunar course, so after a logical interpretation to 11.2 km / s speeds and launched into the cosmos , and only then was the connection with the Luna module ( LM) . This is absolute physical but bullshit ! The coupling of CSM and LM could still be made only at the Earth’s orbit !

2 The first Cosmic velocity of about vB = 7.9 km / s

Now unswervingly on logic and on the physico – mathematical calculations of Apollo 11 to the verification or falsification on the basis of the fuel balance for the start and flight path : To get a body space to the moon, it needs first to the first Cosmic velocity of 7.9 km / s can be achieved , so that you can open out into orbit. In Table 1 , the start and idle masses Mo and ML are the three different stages of the Saturn V rocket addition to actual outflow velocities ve , according to NASA specifications listed (see Table 1).

Table 1 : Start – empty mass and the effective outflow velocities ve ( Leitenberg , B, 2013 in Internet * ) .

Stage N / Fuel Mo ( t) ML (t ) ve (m / s ) Remarks

1 RP (kerosene ) + O2 2286 135 2600

2 H2 + O2 490 39 4200 Is to doubt

3 H2 + O2 119 13 4200 Dto .

CSM +

LM – each Hydra-zin/asymmetrisches Distickstoffte tetroxide and dimethylhydrazine 30 +

15 (Landing

on the moon )

4.9 ( starting from the Moon ) 26

7.2

2.69 2600

2600

2600

Σ

According to NASA, 2930

2940 –

–

– Difference of 10 tons , according to NASA

* Notes to Table 1 : There is the individual parameters very different to the individual authors and some parameters of some authors also ever confused ve example, the specific impulse with the effective exhaust velocity .

According to the Rocket Equation

BB = ve * ln ( Ml + MTr ) : Ml ( 1)

could with the three steps according to a modification of formula (1) theoretically maximum path and speed of the end of firing

BB = 2.6 km / s * ln ( 2930:644 ) + 4.2 km / s * [ ln ( 644:164 ) + ln ( 164:58 ) ] ≈ 2.6 km / s * 1.51 +

* 4.2 km / s (1.37 +1) = 3.9 km / s +4.2 km / s * 2.37 = 3.9 km / s + 9.95 km / s = 13.8 km / s (2)

achieved without taking into account the gravity and air resistance . According to NASA – data from a 440 km orbit height was reached (see also Leitenberg , 2013). So you have the above amount ( 2) according to the formula

DELTA.v = √ 2 * g * H (3 )

with the values used in the Erdgravitationskonstante g = 9.89 m / s ² and orbit height of

H = 440.000 m

DELTA.v = √ 2 * 9.89 m / s ² * 440,000 m = 2.95 km / s ( 4)

essential and immutable deduct from (2). And for the air resistance

P = 0,5 * Fw * v ² * A (5)

1st and 2 Step results by integration of the formula (5) and division by the average of the two masses M1 and M2, the first step , a negative acceleration , and thus a reduction of a velocity generally

DELTA.v = √ 2 * a 2 * H * H * = √ [ (p1 v1 ² * A1 * M1 ) + ( P2 * V2 ² * A2) : M2) ] 6 (6)

where it actually estimated a reduction in the rate of

DELTA.v = √ 2 * 440.000m ² / s ² [ (0.0005 * 4000 * ² 75:1.750.000) + (0.00001 * 10,000 * 75 ² ) : 300000 ]: 6 =

0.3 km / s ( 7)

results . In total it would give a preliminary assessment of

vB = 13.8 km/s- 2.95 km / s – 0.3 km / s = 10.55 km / s (8)

Leitenberg (2013 ) is also only an amount of 10.8 km / s at a height of about 950 km. This must have been used in the bag of tricks of the Apollo repertoire well . Because of the height of 950.000 m would be the difference in altitude of 510 km and an average acceleration of gravity g = 8.5 m / s ² at this level a speed reduction of approximately

DELTA.v = √ 2 * 8.5 m / s ² * 510,000 m = 2.9 km / s ( 9)

been recorded ! This would have an additional mass of fuel

Mtr = ( 2,722,9:4,2 -1) * = 45t ( 2,720,7-1 ) * 45t = (2-1 ) * 45 1 * t = 45 t = 45 t ( 10)

required. Even more precarious sees the overall balance when the burnout rates of 1 (t1 = 161 s ) and the 2 Stage (t2 = 390 s ) for the calculation of the speed reduction approach moves through the Earth’s gravity . Then, would result in a loss of speed even up to an altitude of 188 km according to the formula

DELTA.v = g * (t1 + t2 ) (11)

from

v = 9.89 m / s ² * ( 161s +390 s) = 9.89 m / s ² * 551S = 5.4 km / s (12)

You can twist and turn as you like: Apollo 11 , it was with the CSM and LM Although well into orbit , but never into space with the 2nd Create cosmic speed!

3 The second Cosmic velocity vB = 11.2 km / s

To be able to fly to the moon and other planets , must escape velocity , ie the second Cosmic velocity of about 11.2 km / s can be achieved . Regardless of the illustrations and descriptions of NASA ( ……) the arrangements for the flight to the moon on Apollo 11 to the moon, would have the logic and simple physical considerations, the CSM with the LM with a total mass of M = 45 t 11.2 must km / s embark on the journey to the Moon . So that would be due to the difference in velocity of about 0.4 km / s to Leitenberg (2013 ), an additional amount of fuel according to the mathematically transformed by Rocket Equation ( 1) and conversion to MTr of

MTr = (2.72 0,4:4,2 -1) * 45 t = (2,720,1 -1) * 45 t = ( 1.1 to 1 ) t = 0.1 * 45 * 45 t ≈ 4 , 5 t ( 13)

have been necessary . According to the above calculations of ( 8) 7.2 t even ! It will be used to farther accounting here but the 4.5t .

4 The effect of the gravity of the Earth and moon on the Apollo 11 CSM

To calculate the effect of the gravity of the Earth and moon on the CSM , you have to use the law of gravity . From Newton’s law of gravitation , the following general relation can be derived , the relationship between the two gravitational accelerations g1 ( from the central body ) and g2 ( the spaceship ) and the two radii r1 ( radius of a central body , such as the Earth ) and r2 (distance of the spaceship eg CSM) to reflect a gravitational body :

g2 = g1 * r1 ² (14 )

r2 ²

On a spaceship , for example, the Apollo CSM based at a distance of r 11 of the earth with radius R , can thus be formulated :

g = gE * R ² (15 )

r ²

Now the formula ( 4) must be integrated and divided by r to calculate the average gravitational acceleration can gr . The average size gr gravity is calculated as

r r

g = gE * R ² = ∫ dr 1 gE * R ² | -1 | . (16)

R ² r r r r R

Now the distance from the Earth to be determined by the point r , where the 11 km / escape velocity to be slowed down almost to zero by the action of the average gravitational acceleration gr s . For this, the transformed and Permed formula ( 5)

g = v ² (17 )

2 * r

be equated . It is then , as can be shown

r = 2R * -gE ² . (18)

v ² – gE * 2R

This produces for

r = -2 * 9.89 * 6340000 = 180,657 km ² m . (19)

11,000 ² – 2 * 9.89 * 6340.000

The distance to the moon is that at least another 220,000 km (400,000 km 180,000 km ) . The average positive acceleration to the moon then takes a value according to (5 ) of :

g (220,000 miles) = 1.62 m / s ² * [ ( -1740 ² km ² )) – ( -1740 ² km ²) ] ≈ 0.0078 m / s ² (20 )

* 220,000 * 1740 220,000 km ² 220,000 km ²

to . Thus , the CSM Apollo 11 to the moon to a speed of

v = √ 2 * 220,000,000 m * 0.0078 m / s ² = 2620 m / s ² (21 )

accelerated.

5 The flight of the CSM in lunar orbit , the moon landing and the return flight

To flow into the lunar orbit , had the 2.62 km / s to about 1600 m / s are decelerated . So that would be an additional amount of fuel

MTr = ( 1 :: 2.72 2.6 -1) * 41 t = ( 2,720,38-1 ) * 41 t = (1.46 -1) * 41 t = 0.46 * 41 t = 18 , 89 t ( 22)

been necessary . For landing on the Earth’s natural satellite of the Mondortbit would be taking into account the moon’s gravity , resulting in additional speed to the lunar surface of around 402 m / s is produced (v = √ 2 * 100,000 * 1.62 = 402m / s) 8.4 more t of fuel was required , it can be shown as follows impressively :

MTr = ( 2.722 : 2,6 -1) * 7.2 t = (2,720,77 -1) * 7.2 t = (2.16 -1) * 1.16 * 7.2 = 7.2 t t t = 8.4 . (23)

Thus, the Lunar Module ” Eagle” would have already been fired over their powder , to put it bluntly , because it began with a total of only 7.8 tonnes of fuel available. This is again a negative fuel balance , in this case from 0.6 t = 8.4 t 7.8 t before . For the start of the lunar surface in the lunar orbit would have a further amount of fuel

MTr = ( 2.722 : 2.6) -1) * 2.7 t = (2,720,77 -1) * 2.7 t = (2.16 -1) * 2.7 * 2.7 t = 1.16 t t = 3.13 ( 24)

made necessary. Thus missing a total of about 3.7 tonnes of fuel to land with the Lunar Module “Eagle ” on the moon and return from there again . Furthermore, other 9.4 tonnes of fuel would have been required to get out of the gravitational field of the moon. The 9.4 t calculated according to the escape velocity of 2.3 km / s from the moon :

MTr = (2.72 0,7:2,6 -1) * 30 t = (2,720,27 -1) * 30 t = ( 1.31 to 1 ) * t 30 = 0.31 * 30 t = 9 , 4 t . (25)

6 The return flight to Earth

The 2.3 km / s line speed will be slowed down again generated by the moon’s gravity negative acceleration to zero. This distance from the moon errechtet according to ( 7) and similar to ( 8) as follows

r = -2 * 1.62 * 1740,000 m = 28,220 km ² . (26)

2300 ² – 2 * 1.62 * 1740.000

To the ground accordingly remain still considered reciprocally 372,000 km . The average positive acceleration is calculated according to the soil (5)

g (3720.000 km ) = 9.89 m / s ² * [ ( -6340 ² km ² )) – ( -6340 ² km ²) ] ≈ 0.165 m / s ² . (27)

* 372,000 * 6340 372,000 km ² 372,000 km ²

Thus, the CSM is from Apollo 11 to a speed to Earth

v = √ 2 * m * 0.165 372 000 000 m / s ² = 11.078 km / s ( 28)

accelerated. The approximately 11.1 km / s , however, must be slowed down to about 8 km / s orbital velocity again to get into orbit. This would be a further quantity of fuel

MTr = (2.72 3:2,6 -1) * 30 = 2.72 1.2 t -1 ) * 30 t = ( 3.3 to 1 ) * 30 t = 30 t = 2.3 * 69 t ( 29)

have been necessary . For the specious argument that the CSM was decelerated supposedly of 11 km / s directly to 0 km / s in Earth’s atmosphere can , so do not apply , because instead of the kinetic energy of about 0.9 TJ (with v = 8 km / s ) 1.8 TJ (at 11 km / s) into thermal energy would have to be converted . Thus, the immersion temperature would have increased the Earth’s atmosphere at about twice that can be shown. Because if one equates the kinetic energy of the thermal energy , then:

Ekin = Etherm = 0.5 * m * v ² = m * T * R. (30)

Thus, first, the mass m out once shortened may be formulated and

0.5 * v ² = T * R. (31 )

This results in relation to v2 from v1 = 11 km / s = 8 km / s, a ratio of the immersion temperatures of

² v1 : v2 ² = T1: T2 = 121:64 ≈ 2:1 ( 32)

( R is the gas constant and the factor 0.5 is also cut out after relations education ) . To demonstrate this relation with a concrete example , is appended hereto , that the heat protection tiles of the space shuttle as it enters the Earth’s atmosphere at an inlet velocity of 7.6 km / s at least 1600 K were generated (see various authors on the Internet, 2013 ) . According to the above formula ( 28) would then have an entry speed of at 11.2 km / s at a temperature of

11.2 ² * 1600 K: 7.6 ² ≈ 3475 K (33 )

the command module of Apollo upon entering Earth’s atmosphere must be created in serving on heat shield . Apollo 11 and N would therefore like a shooting star in Earth’s atmosphere by the contemporary state of technology and technology burns !

7 conclusions

Thus, there is an additional amount of fuel in the sum of at least

MTr = 18.89 t + 4,5 t + 8,4 t + 3.13 t + 9,4 t + 69 t -7.8 t -4, t = 102.52 t ≈ 102 t . (34)

And this mass is quite remarkable when one considers that the third Stage of the Saturn rocket possessed only about 100 tonnes of fuel . On the other hand, stood there in total , according to data from NASA Earth orbit only 4 +7.8 t t = 11.8 t available to cope with the moon project! In other words, Apollo 11 and Apollo N may have never happened !

Epilogue: The question is permissible , as the Americans should have mastered in the summer of 1969, the Apollo 11 project at all? How they staged this, it is purely just been physically and mathematically impossible! The former three American astronauts will have probably found eight days in orbit , as has been impressively documented the early eighties with a movie and then return back to Earth. Apollo 11 was the biggest bluff of human history and great cinematically staged, with many errors !

8. Epilog

8.1.Flight time , speed, and cosmic radiation

In some films it is claimed that the astronauts with their spaceship within just 3 days (69 h to 72 h – now what 69 or 72 h) would have reached the moon. This is first of all not at all, because according to purely formal physical-mathematical calculations at least 87 h at an average acceleration of g = 0.0078 m / s ² one (see below formula 20 , which was calculated gm ) from the Earth to the Moon at a distance of 380,000 kilometers needed! After all, the time t is calculated as t = √ 2 * s: gm = √ 2 * 380,000,000 m: 0.0078 s ² ≈ 87 h This Apollo 11 was already disproved by an ” own goal ” of the TV shows !

With the moon (-) and the stars ( railways ) but it is such a thing ! After Sternfeld (1959 ) only two 15 -day constellation and a 60-day Szenarion should exist to reach the moon by an artificial spacecraft from Earth and land on the ground again . Regardless of the theoretical facts and details of star field , the required research satellite SMART I, which was launched in late September 2003 , 49 days until on moon level and five months until he einmündete in lunar orbit . And the latest lunar expedition of Chinese spacecraft Chang ` e-3 in December last year covered the distance from Earth to the moon within 15 days back . This Apollo 11 would also astrophysical impressive empirically refuted because a rogue 8-day regimen that is allegedly practiced astrophysical does not exist !

And an astronaut stiffened to assert that they dashed at 8 km / s through space . Again, this is purely physical terms, impossible! Because: The average velocity is given by v = √ 2 * s * gm = √ 2 * 380,000,000 * 0.0078 m / s ² = 2434.7 m / s ≈ 2.4 km / s This was the second own goal !

8.2. The cosmic radiation that would have influenced the astronauts would have been just hopeless ! The astronauts would board the flight back simply did not survive to the moon and because they had been exposed unprotected to a radiation dose of 768 Sievert and at an absorption rate of 90 percent , which must be referred to simply as utopian, the radiation dose would still be almost 77 sievert. Because: The particle is 1000 elementary particles per second and square meter [ about the surface of the body , see human A. Sternfeld (1959 ) : Artificial Satellites , B * G * TEUBER PUBLISHING COMPANY * LEIPZIG , 1959, and Lindner , 1966 ] outside the magnetic field of the earth ( approximately up to 45,000 km). Calculated on eight days lunar mission , the number of protons would ( at 85 percent of the total radiation by Sternfeld , 1959) , take the one astronaut would

N = 691,200 s * 0.85 * 1000 * 1 / s ≈ 0.6 * 109 ( 1)

^

amount (8 d = 8 * 24 * 3600 s = 691,200 s ) . A proton has the energy of

EProton = 0.6 * 1015 eV (electron volts – see Sternfeld , 1959, Lindner, 1966 ( 2)

and Internet 2009).

This results in a total amount of energy of

EΣ = 0.6 * 109 * 0.6 * 0.36 * 1015eV = 1024eV . (3)

A eV represents the amount of energy of 1.6 * 10 – 19J .

Thus the total energy in Joules is calculated

EΣ = 0.36 * 1024 * 1.6 * 10 – 19J = 0.576 * 105 = 57600 J. ( 4)

Based on an average body weight of 75 kg , you have to go to the unit of radiation dose in sievert ( Sv) , divide the 57600 J by 75 kg and then receives 768 J / kg and thus a radiation dose of approximately 768 Sievert (1J / kg = 1 Sievert ) . For comparison, a result of the atomic bomb on Hiroshima died all those affected in the aftermath , which were exposed to radiation doses of 6 Sv ! And at 10 Sv one is killed on the spot In other words : The American astronauts were arrived as corpses on the ground. This is negated impressively on a third physical layer Apollo 11 to N!

Siegfried Marquardt, Kingswells, the 17/01/2014

Siegfried Marquardt, on January 31, 2014 at 10:49 pm said:Refutation of Apollo 11 on the basis of physical evidence at four levels

1 After Sternfeld (1959 ) only two 15 -day constellation and a 60-day Szenarion should exist to reach the moon by an artificial spacecraft from Earth and land on the ground again . Regardless of the theoretical facts and details of star field , the required research satellite SMART I, which was launched in late September 2003 , 49 days until the moon level and five months until he einmündete in lunar orbit . And the exfolgreich running in the December 2013 lunar expedition of Chinese probe Chang ‘ e-3 proved in 2013 captivatingly that one needs at least 14 days to deal with the distance to the moon. This Apollo 11 would already disproved empirically , because a rogue 8-day regimen that is allegedly practiced astrophysical does not exist !

2 The 800- Sv to cosmic radiation , which would have influenced the astronauts within the eight days would easily have been hopeless ! Because: This is the 80 -fold lethal dose of radiation. The astronauts would board the flight back simply did not survive to the moon and .

3 It was missing a total of over 100 tons of rocket fuel to get from Earth to the moon and from there back to Earth at the predetermined by NASA loop-shaped trajectory . Furthermore, the amount of fuel and the former fuel parameters would have a lunar cargo and even boot from the moon under the former conditions impossible.

4 Already in a preliminary phase of the reconstruction of the land on the Moon ferry after deducting the alleged approx MTr = 10.6 t invoiced fuel mass from the starting mass Mo = 15 t of the Lunar Module merely remain only 4.4 tonnes of empty weight , which already in the material reconstruction the cabin (approx. 1.1 tons ) , parts of the outer cell ( about 1.3 t) , the support legs (about 0.7 t) and the load ( approximately 1.7 tons ) , excluding the weight of the astronauts and their space suits (400 kg), the mass of the tanks and the two main engines of the lunar module (…) with nearly 400 kg exceeded.

Overall, missing nearly 3 tonnes mass , such as 11 could be impressively demonstrated with the total reconstruction of the lunar module of Apollo.

Siegfried Marquardt Kingswells, the 31.01.2014

phoenixpics, on February 1, 2014 at 11:07 am said:You know, I am so weary and sick of this bullshit. Haven’t you nutters got better things to do?

Siegfried Marquardt, on February 28, 2014 at 10:19 pm said:Re- and deconstruction of the command module of Apollo 11

The command module (CM) of Apollo 11 is according to Internet information (see Wikipedia , February 27, 2014) a mass of m = 5.9 t , a height of H = 3.23 m, a diameter of d = 3.9 m and an internal volume of Vi have possessed = 6.17 m³. The total Vtotal would therefore

Vtotal = d ² * π * H : (3 * 4) = 3.9 ² m² * 3.14 * 3.23 m: 12 = 12.86 cubic meters ( 1)

must be . Thus, the volume would have Vzelle around for the wall of the cell of the CM

Vzelle = Vtotal -Vi = 6.71 m³ ( 2)

must assume . This raises the legitimate question of what material the command module was manufactured ? The command module will surely have passed not out of cardboard or even paper , because the wall of the cell would have a density widely accepted less than 1 if one forms the quotient of mass and volume . The density ς is namely using the figures used and calculations made to the volumes

ς = 5.9 t : 6.71 m³ = 0.879 kg / dm ³. (3)

This could not even afford paper and cardboard . Well the outer wall of the CM could have consisted of aluminum. At a density of ς = 2.7 t / m³ would at a mass of 5.9 t an available volume to

V = m: ς = 5.9 m³ : 2.7 = 2.2 m³ ( 4)

result . To determine the wall thickness of the cell , the inside diameter di of the cubic equation must

0 = di ³ – (H -da ) * di ² – since ² * H – (V * 12: π ) (5)

be calculated , which can be derived from the calculation formula of a double cone. If the values for the diameter da = 3.9 m, for the height H = 3.23 m and uses available for the volume of V = 2.2 m³ in the above equation of the third degree , then results

46.53 +0.67 * di ² – ³ di = 0 (6)

The solution of this cubic equation for the internal diameter di is then about di = 3.85 m . Thus, the command module would be with an outer skin of 50 mm : 2 = 2.5 cm raced through space ! But logically, would still be missing the heat shield to ensure a safe landing on Earth. Assuming that half of the mass of the cell had been made for the heat shield is available, then the cell would have only an outer skin of about 1.3 cm. In this case , however, the heat shield would only have a wall thickness of 2 mm can have steel , as can be calculated in an analogous manner with the formula (139). A comment is not necessary to almost completely : Apollo 11 would burn up like a shooting star in the atmosphere ! In other words, even the CM was the figment of a bad design – just a chimera ! The American engineers and American astronauts were really true heroes !

Siegfried Marquardt, Kingswells, the 28.02.2014

Siegfried Marquardt, on April 5, 2014 at 8:56 am said:6 In addition, the pendulum behavior of the flag on the moon is extremely treacherous ! Because the pendulum period T , which is physically connected to the pendulum length l (l = 0.7 m) and the acceleration of gravity g ( g = 9.81 ) to

T = 2 * π * √ l : g (1 )

calculated , would have on the moon

T = 6.28 * √ 0.7 m: 1.6 m / s ² ≈ 4.2 s ( 2)

amount . In the TV film documentaries the period is about 2 seconds but , as indicated on the earth. The exact calculation of the period of the earth results in precise

T = 6.28 * √ 0,7 m / 9.81 ≈ 1.7 s ( 3)

This difference in time of 2.5 s is serious and would in fact be for the layman in the TV documentaries in the consecration of the flag on the “moon ” perceptible. Furthermore, it should on the moon , a slightly damped periodic oscillation arise because no atmosphere is present on the moon. The true vibration is increasing but almost aperiodic and the period T is only about 1.7 s Summary: The shooting took place on Earth !

7 One graphic by Stemmer (aerodynamics of spacecraft re-entry aerodynamics, TUM , Munich , 2012) could be seen that in the first few minutes at a speed of 8 km / s on the heat shield of spacecraft a temperature of about 3100 K occurs. This is also the reason that during re-entry of spacecraft , the first Cosmic velocity is reduced by severe braking thrusters. Then s would have to rise to a soak temperature of 6121 K at the heat shield of the Apollo command module as it enters the Earth’s atmosphere at an inlet velocity of nearly 11.1 km /s . ” Space capsule CM Columbia ” ( http://www.bredow – web.de, 2013) could be seen that the heat shield of the command module CM was supposedly designed for only 2726 K to the web document – this could no longer save the CM Apollo 11 !

In other words, Apollo 11 and N s would be like shooting stars burns at a dipping speed of 11,1 km /s in the atmosphere according to the state of that technology and engineering !

Siegfried Marquardt, Kingswells, the 05/04/2014

Siegfried Marquardt, on December 28, 2014 at 6:56 pm said:Abstract mathematical-physical refutation of Apollo 11 and N

1. After Sternfeld (1959) only two 14-day constellations and a 60-day scenario should exist to reach the moon with an artificial spacecraft from Earth and land on the earth. Regardless of the theoretical facts and details of Sternfeld, required the research satellite SMART I, which was launched end of September 2003, 49 days until the moon level and five months until the probe einmündete in lunar orbit. And successfully running in the December 2013 lunar expedition of Chinese probe Chang`e-3 proved impressively that it takes at least 14 days to cope with the distance from the Earth to the Moon. This Apollo 11 would already impressively refuted empirically because a putative 8-day regime that is allegedly practiced with Apollo 11 and drilled, astrophysical theoretically and empirically does not exist!

2. The cosmic radiation, which would have affected the astronauts within eight days would have been absolutely hopeless! After all, you would have incorporated a lethal dose of at least 11 Sv to 26 Sv depending on the chosen model calculation. if you are in this context to the high-energy particle density in the cosmos and to the particle stream the sun with the solar constant of 8.5 * 1015 MeV / m * s thinking. The astronauts had the flight to the moon and back in any case not survive.

3. It was missing a total of 163 tons of rocket fuel to get from Earth to the moon and from there back to Earth by NASA on the given loop-shaped trajectory. Furthermore, the amount of fuel and the former fuel parameters would have a moon charge and even boot from the moon under the former conditions impossible. Alone for the transition from the elliptical trajectory close to the Moon would be for the braking of the CSM + LM with a total of 45.3 t mass of the 2.3 km / s to 1.7 km / s for the lunar orbit [2,72 high (0 , 6: 2,6) -1] * t = 45.3 (1.26 -1) * 45.3 * 45.3 t = 0.26 ≈ 12 tonnes of fuel have been necessary! The remaining three tons a moon landing would not have been possible and start from as little moon! On the Moon, LM did not have 14 t, but (15-8) t = 7 t!

4. Reconstruction of the command module at a predetermined height by NASA of 3.23 m and a diameter of 3.9 m, resulting in the end can only result a total volume of about 12.9 m³, showed that after deduction of the declared internal volume of 6.23 m³ volume of the outer cell of the command module only about 6.7 m³ could include. With a mass of 5.9 t the density of the command module would thus have to be only about 0.9. This would “afford” not even paper or cardboard! Another mathematical optimization was then that the outer cell only from a 2.5 cm thick aluminum layer could exist – without the heat shield. If one half of the total mass of 5.9 tonnes for a heat shield as a basis, the heat shield could consist of only 2 mm thick steel. A commentary is superfluous almost: The command module would be in the earth’s atmosphere with a theoretically calculated braking temperature of at least 45,000 K like a shooting star burns!

5. Even in a preliminary phase in the reconstruction of the Lunar Module according to NASA parameters after deduction of the alleged approx MTr = 10.8 t invoiced fuel mass of the starting compound with Mo = 15 t the Lunar Module merely remain only 4.2 t to empty weight, already with the material reconstruction of the cabin (about 1.1 tons), parts of the outer cell (1.3 t), and the declared weight (1.7 t), without taking into account the weight the astronauts with their space suits (400 kg), the mass of the tank and the two main engines of the Lunar Module (…) of 600 kg exceeded. Total lacked exceeding 3 tonnes construction mass, could be as originally stated by NASA and how 11 is impressive and convincing with the total reconstruction of the Luna module of Apollo.

6. The declared by NASA thrust of 44.4 kN and 15.6 kN of the descending and ascending level does not match with the theoretically calculated thrust. There are significant differences here! (descending level: S = m * ve = 16.8 kg / s * 2560m / s ≈ 43 kN and rising level: S = 5.9 kg / s * 2560m / s = 15.1 kN).

7. addition would be the Lunar Module at a speed of 215 m / s bounced and crashed on the moon, because the former fuel parameters such as the effective exhaust velocity of 2560 m / s and the mass ratio of the descending level of 15 t to 6.8 t only permitted a maximum speed burnout of 2025 m / s [vB = ve * ln (Mo: ML) = 2560m / s * ln (15: 6,8) = 2560m / s * 0.79 = 2025 m / s]. Taking away the 570 m / s, which are caused by the moon’s gravity from, so you get only a resultant velocity of 1455 m / s. It could have been so ago by the technical and physical parameters, can be held no moon landing!

It is on the other hand almost pointless to be mentioned that the rising level only a resulting burnout velocity of around 1500 m / s could have and therefore does not enter the orbit would be as it had a speed difference to the orbital velocity of 170 m / s in this case would.

8. Furthermore, the pendulum behavior of the flag on the moon is extremely treacherous! For the pendulum period T, which is physically connected to the pendulum length l (l = 0.7 m) and the gravitational acceleration g (g = 9.81) to

T = 2 * π * √ l: g (1)

calculated, would have on the Moon

T = 6.28 * √ 0.7 m 1.6 m /s ≈ 4.2 a (2)

respectively. In the TV film documentaries period lasts but close to 2 s, as indicated on the earth. The exact calculation of the period for the earth yields accurate

T = 6.28 * √ 0.7 m / 9.81 ≈ 1.7 s. (3)

This time difference of 2.5 s is serious! In addition, a slightly damped periodic oscillation would arise on the moon, because there is no atmosphere is present on the moon. The increasing vibration is true but almost aperiodic. Summary: The shooting took place so unique on earth!

9. A mechanical instability of the lunar module would have made an intact moon landing impossible! Every person on the planet has probably already seen a failed rocket launch when the rocket has already picked up a few meters from the launch pad and then fail the engines and do not produce more power. As a result, the rocket moves the physical laws of gravity accordingly again towards the launch platform and then tilts due to the mechanical instability simply because the center of gravity has changed dramatically. This would also be the fate of the lunar module of Apollo 11 was because shortly before landing an absolute instability of the ferry would have been! Because: Full expected gross, the rising level would have had to ground just before landing on the moon for about 5 t and the descending stage would have received under the fuel consumption of only 8 t only about 2 tons of empty weight had. As the focus of the Lunar Module must have lain on the moon exactly at 2.1 m before landing the ferry across the nozzle, the torques would like 2.5: 1 to 3:1 behaved. For an absolutely unstable mechanical system would be active! Even the smallest vibration, such as vibrations through the engine orpressure fluctuations in the effluent gases in the nozzle of the engine have the lunar position ferry can easily tip over!Amoon landing would indeed be “successful”, but a return from the moon would have been so impossible. However, sincê 11 have fortunately survived the imaginary adventures all actors of Apollo, it can be concluded razor sharp, no moon landing took place.

The solution of the physical problem is that the focus of a lander simply must be at the level of the nozzle of the engine, such as the Chinese realize this in December 2013, and practiced.

P. S. By the way, the author had the skeptical thoughts on the instability of the lunar module landing on the moon more than 45 years ago spontaneously for about 1 s had entertained!

Siegfried Marquardt, Kingswells, December 2014

Siegfried Marquardt, on May 27, 2016 at 7:59 pm said:Analysis of the film Apollo 13, which was broadcast by the TV station Vox on 16/05/2016

In the film, voiced the protagonist of Apollo 13 the following comments, the absolutely refute Apollo 13 and the other Apollo missions:

1. Multiple layer aluminum foil only separate us from the universe meant an astronaut. The CSM would simply explode in this physical structure, because at an internal pressure of 1 bar on the outer shell of the CSM, a force of F = po * AM = 1 kgf / cm² * 10,000 * 3.14 * 3√3,2²m² + 1, 9²m² = 10,000 * 9.3 * 3.7 kp = 344,000 kp = 344 Mp = 344 tonnes would have looked!

2. the increase in CO2 concentration was maintained at 15 percent and is constantly displayed on the instrument. A CO2 concentration of 15 percent mean 300 g / m³ CO2 [4 * 2 * kg / m³ * 15: (0.04 * 10000) = 300 g / m³]. The threshold for CO2 is 9 g / m³ and would have been exceeded by the more than 33- fold. The astronauts would be smothered in no time and have been dead!

3. It should be the electric current is reversed! This is physical nonsense

4. To study the negative thrust are turned on! This is absolutely absurd! (The lunar module was behind the moon load module, as could be seen).

5. The expected ballast for the lunar rocks should be compensated! This is physical nonsense!

6. The reverse thrust should be switched on! This is technical-physical bullshit!

7. The heat shield with the entry into the atmosphere should be heated to 2000-2700 degrees. The real value was at 8 km / s over 8600 degrees Celsius under the existing conditions. In 11 km / s, the heating is even 16,000 degrees.

8. The outside temperature load should allegedly be -187 degrees Celsius. Correctly are -273 degrees in space!

9. The entry speed in the atmosphere should have = 32,000 ft / s 9.6 km / s. The true value is 11.2 km / s, as they supposedly came from the moon.

For Apollo 11 would be to N disproved! Because the Americans did not even know the correct physical parameters for a space flight to the moon.

Siegfried Marquardt, Koenigs Wusterhausen

phoenixpics, on June 29, 2016 at 10:38 am said:Bored now. Go away.

Siegfried Marquardt, on June 22, 2016 at 5:50 pm said:NASA refutes itself with documentary on Apollo 13

In this film documentary on Apollo 13 on 06.11.2016 by the TV station N 24, the alleged near disaster was discussed, where supposedly the oxygen supply in the Command Module CM collapsed. Then the astronauts boarded the lunar module as a lifeboat in fact. The flight director of NASA pondered then abort the flight to the moon and to reverse Apollo 13 directly to earth. How should it work astrophysical mean? In the best case would be to assuming that Apollo 13 was at the level of the neutral point of gravity of the Earth and Moon on the way to the moon, an amount of fuel with the specified by NASA Fuel combination of hydrazine / dimethylhydrazine as fuel and nitrogen tetroxide (N2O4) as the oxidizer with an effective exhaust velocity ve of about 2.6 km / s of

MTr = [1- (1: 2.72 11.31 2.6] * 43,7 t ≈ 43.1 t (1)

have been necessary! Thus, NASA itself has refuted because this amount of fuel at all was not available!

Siegfried Marquardt, Koenigs Wusterhausen

phoenixpics, on June 29, 2016 at 10:35 am said:You’re talking absolute bollocks, again. Please stop wasting your time, and mine, and post your woo-woo conspiracy theory scientifically-ignorant bullshit on a site that shares your ridiculous views.

Siegfried Marquardt, on July 1, 2016 at 5:12 pm said:The analemma of Apollo 11’s astrophysical nonsense!

The propagated by NASA and declared eight loop of Apollo 11 to the moon and back to Earth is simple astrophysical nonsense because moving planet, satellite planets and spacecraft after the First Kepler’s law in elliptical orbits around the central stars, planets and satellite! With the insane declared by NASA rollercoaster of Apollo 11, the energy or the fuel consumption would have increased by several times. The resultant velocity vr to the confluence with the lunar orbit and return and the confluence into orbit would thus generally to about

vr = √vo² + 2 * VO = √3 * VO ≈ 1.73 * vo (1)

increased, vo is the orbital velocity of the moon and Earth orbit. Thus, the fuel consumption increases as the mouth in the moon and orbit in general on

MTr = [1- (1: 2,72vo * 0.73: ve)] * mo. (2)

For the mouth into the lunar orbit by itself, fuel consumption is thus calculated on

MTr = [1- (1: 2,721,24: 2.6)] * 43,7 t ≈ 17 t. (3)

For the moon landing a fuel composition gives to

MTr = [1- (1: 2,722,2: 2.6)] * 15 t ≈ 8.6 t (4)

and for the re-entrance into the orbit results in a fuel mass

MTr = [1- (1: 2,722,2: 2.6)] * 4.7 t ≈ 2.7 t (5)

Returned from the moon to reach the escape velocity would be a fuel composition of

MTr = [1- (1: 2,721,24: 2.6)] * 17 t ≈ 6 t. (6)

For Apollo 11 would have been absolutely Missed his powder because only 18.5 t (Service Module) and 10.8 t were for the lunar lander total available.

Siegfried Marquardt, Koenigs Wusterhausen

The analemma of Apollo 11’s astrophysical nonsense!

The propagated by NASA and declared eight loop of Apollo 11 to the moon and back to Earth is simple astrophysical nonsense because moving planet, satellite planets and spacecraft after the First Kepler’s law in elliptical orbits around the central stars, planets and satellite! With the insane declared by NASA rollercoaster of Apollo 11, the energy or the fuel consumption would have increased by several times. The resultant velocity vr to the confluence with the lunar orbit and return and the confluence into orbit would thus generally to about

vr = √vo² + 2 * VO = √3 * VO ≈ 1.73 * vo (1)

increased, vo is the orbital velocity of the moon and Earth orbit. Thus, the fuel consumption increases as the mouth in the moon and orbit in general on

MTr = [1- (1: 2,72vo * 0.73: ve)] * mo. (2)

For the mouth into the lunar orbit by itself, fuel consumption is thus calculated on

MTr = [1- (1: 2,721,24: 2.6)] * 43,7 t ≈ 17 t. (3)

For the moon landing a fuel composition gives to

MTr = [1- (1: 2,722,2: 2.6)] * 15 t ≈ 8.6 t (4)

and for the re-entrance into the orbit results in a fuel mass

MTr = [1- (1: 2,722,2: 2.6)] * 4.7 t ≈ 2.7 t (5)

Returned from the moon to reach the escape velocity would be a fuel composition of

MTr = [1- (1: 2,721,24: 2.6)] * 17 t ≈ 6 t. (6)

For Apollo 11 would have been absolutely Missed his powder because only 18.5 t (Service Module) and 10.8 t were for the lunar lander total available.

Siegfried Marquardt, Koenigs Wusterhausen