So, who wants to win 1 million Euro?

[Count Zero]

At departure Earth the Command Module and the Service Module are together loaded on top of the Saturn rocket with the Lunar Module stored below the Service Module, actually below the rocket engine outlet of the Service Module.

After lift off and one orbit Earth the space ship is sent off towards the Moon and one way or another the Lunar Module is shifted to the top of the Command Module, so that later, in Moon orbit, two persons can enter it via the hatches. Can anybody explain how the transfer of the Lunar Module from below the Service Module to the top of the Command Module was done?

Cripes!  He doesn't even know about the transposition, docking & ejection maneuver (which was illustrated in everything from the original press kits to a major a major motion picture, and was televised live during the missions) and he wants to lecture us on Apollo?! 

He probably senses he's getting into trouble with his fake rocket equations (the bluff isn't working), and is trying to change the subject.  Let the Gish-Gallop begin...

[Chew]

Please explain how you got this 75.47 GJ result:

To reduce the speed of a mass of 43 000 kg from 2 400 to 1 500 m/s you need 75.47 GJ brake energy! If 1 kg rocket fuel produce 1.63 MJ energy it seems you need 46 300 kg fuel for this maneouvre. You should wonder, where it was carried.

43000*(2400²-1500²)/2

It is basic physics. See table at end of article in link given in post #1, where all is explained.

First of all, that is the wrong equation. It takes x amount of fuel to accelerate by 900 m/s in free fall (neglecting relativistic effects) regardless of your initial speed, whether it be 0 or 100,000 m/s. According to that equation, it will take 23 times as much fuel to accelerate from 10,000 to 10,900 m/s than from 0 to 900 m/s and that is just wrong.

Second, your energy density is for hydrazine used as a monopropellant. The SPS used Aerozine 50 as the fuel and nitrogen tetroxide as the oxidizer. In rocket propulsion, the proper units you want to use are specific impulse or exhaust velocity. Aerozine 50/N₂O₄ had about a 50% higher I_sp.

Sorry, you are wrong. To start with you have to decelerate, i.e. slow down at arrival Moon

To emphasize "deceleration" is just another one of your blunders that proves you are not an engineer. A real engineer would know that "acceleration" simply means a change in velocity. Everything else you said is just as wrong as the first time you said it. Learn the difference force and energy.

[nomuse]

I am evidently comparing the kinetic energy of the mass at two different speeds, 2400 and 1500 m/s, so my formula and calculations are correct.

Why?  Why wouldn't you just use the Ke of the difference?  Your method gives different answers for the same change in velocity, depending on starting conditions!

[Chew]

[nomuse]

?? So how was the Lunar Module shifted from below the Service Module rocket outlet to the top of the Service Module in space?
Did Walter Cronkite do it? How? How was the Lunar Module actually connected to the Service Module at departure (below the Service Module rocket engine outlet)? And how was the Lunar Module disconnected and then shifted to the top of the Command Module? Any link to NASA reports about that?

Why?

How is this an impossible task?  What makes it different from orbital rendezvous (which goes way back to Gemini) -- except that the later is TOUGHER.  Why the sudden fixation on one detail out of hundreds of thousands?  Why is it you don't know anything about a basic and necessary part of the mission profile?  How is it you think you can calculate the cost of any of the major burns without knowing what kind of spacecraft had to make them (aka short stack, S-IVb on or off, etc.)

[Heiwa]

The kinetic energy of a mass m = 43000 kg at velcocity v=2400 m/s is evidently 43 000*2400²/2= 123.84 GJ
At v=1500 m/s the kinetic energy is 48.375 GJ.
The difference in kinetic energy of a mass of 43000 kg at 2400 and 1500 m/s is therefore 123.84-48.375=75.465 GJ.
In order to reduce the velocity from 2400 to 1500 m/s, which takes a certain time t (seconds) you must apply a force F (Newton), while the space ship displaces a distance d (meter).
Say that the time t is 600 seconds? What is the force F? And the distance d? Show me that you can calculate.

[nomuse]

If I'm remembering any of my numbers right, the RCS had a delta-v capacity some 1,000x larger than the kind of velocities called for in the re-arrangement of the vehicles. 

[Heiwa]

Thanks for the photo of the Service and Command Modules together without any Lunar Module attached.

The Lunar module was apparently fitted below the Service Module rocket engine outlet at departure Earth and later, by somebody called Walter, shifted to the top of the Command Module in space. Can you please explain how it was done, e.g. by some photos and links to a suitable NASA report.

[Andromeda]

Thanks for the photo of the Service and Command Modules together without any Lunar Module attached.

The Lunar module was apparently fitted below the Service Module rocket engine outlet at departure Earth and later, by somebody called Walter, shifted to the top of the Command Module in space. Can you please explain how it was done, e.g. by some photos and links to a suitable NASA report.

You cannot be for real.

[Heiwa]

Of course I am real! Like my €1 000 000:- at my bank.

Here we see the Lunar Module connected to the top of the Command Module in space prior ariival Moon. But at departure Earth the Lunar Module was connected to the Service Module below the rocket engine outlet.
How was the transfer of the Lunar Module done?
How did Walter do it?

[nomuse]

The kinetic energy of a mass m = 43000 kg at velcocity v=2400 m/s is evidently 43 000*2400²/2= 123.84 GJ
At v=1500 m/s the kinetic energy is 48.375 GJ.
The difference in kinetic energy of a mass of 43000 kg at 2400 and 1500 m/s is therefore 123.84-48.375=75.465 GJ.
In order to reduce the velocity from 2400 to 1500 m/s, which takes a certain time t (seconds) you must apply a force F (Newton), while the space ship displaces a distance d (meter).
Say that the time t is 600 seconds? What is the force F? And the distance d? Show me that you can calculate.

Why are you working backwards?  If you were calculating an elastic collision, would you convert to Ke before working your way back to MV?

Anyhow...here's a fun;  2,400 m/s - 1,500 m/s = 900.  1/2 43,000 kg (21,500)  x 900^2 (810,000) = 1.74 x 10^10 joules.  Odd, isn't it, how the base velocity seems to matter in your version. 

[Andromeda]

Of course I am real! Like my €1 000 000:- at my bank.

You have been asked for proof of that and ignored the request.  Why?

Here we see the Lunar Module connected to the top of the Command Module in space prior ariival Moon. But at departure Earth the Lunar Module was connected to the Service Module below the rocket engine outlet.
How was the transfer of the Lunar Module done?
How did Walter do it?

[Heiwa]

Why is it you don't know anything about a basic and necessary part of the mission profile?  How is it you think you can calculate the cost of any of the major burns without knowing what kind of spacecraft had to make them (aka short stack, S-IVb on or off, etc.)

Space travel is similar to a voyage at sea (my speciality) and you need fuel to get from A to B. If you study my presentation (topic - see post #1 with link to it) you see my concern is just the fuel used by Apollo 11 and ... after basic calculations using NASA input ... I find that Apollo 11 could not carry the required fuel to get in and out of Moon orbit. The space ship was too heavy or the engines to inefficient or something.
If you think my calculations are wrong, just show it. Do not tell me how stupid I am, etc, etc.

I am also curious to know how Walter managed to shift the Lunar Module in space from one end to the other of the CSM! Do you know, how Walter did it?

[Andromeda]

Space travel is similar to a voyage at sea (my speciality)

Let me stop you right there.  Space travel is in three dimensions with hugely varying mass and changing gravitational fields.  Neither of those apply to sailing.

If you think my calculations are wrong, just show it.

We did.  You ignored us.

I am also curious to know how Walter managed to shift the Lunar Module in space from one end to the other of the CSM! Do you know, how Walter did it?

I showed you that - look at the diagram I provided in post 71 of this thread.

[Heiwa]

Of course I am real! Like my €1 000 000:- at my bank.

You have been asked for proof of that and ignored the request.  Why?

Here we see the Lunar Module connected to the top of the Command Module in space prior ariival Moon. But at departure Earth the Lunar Module was connected to the Service Module below the rocket engine outlet.
How was the transfer of the Lunar Module done?
How did Walter do it?

So in Earth orbit the CSM was disconnected from the last stage of the Saturn rocket, rotated 180° and then connected to the Lunar Module? And Walter did it? And then the CSM with the Lunar Module on top of the CM was sent off to the Moon. Why not?

Question remains how they managed to brake to get into Moon orbit three days later.