So, who wants to win 1 million Euro?

[JayUtah]

2400 m/s is the arrival speed at the Moon according NASA.

"Arrival" is not a recognized technical term in astrodynamics.  Please use appropriate terminology, and specify at what point along the orbit this "speed" occurs.

1500 m/s is the speed in orbit around the Moon according NASA.

The initial orbit after LOI-1 and before LOI-2 is elliptical, not circular.  Please give more specific information about the orbit you envision the spacecraft to have entered.

Also, "according to NASA" is rhetorical nonsense, since the principles of orbital mechanics predate NASA by 100 years or more and are not dictated arbitrarily by them.  The Apollo orbits are dictated by this century-old science and derived according to it.  No need to try to accuse NASA at every turn.

43000 kg is the mass of the space ship at arrival according NASA.

Evidently it changes when fuel is consumed - but I keep it constant...

Then you did the problem spectacularly wrong.  You cannot pretend using the wrong propulsion model will give you the right answer.  The mathematical management of the variable-mass property of a space vehicle is what separates the real engineer from the incompetent amateur.  A real engineer can derive the mass changes based on the known properties of the propulsion system, or the change in the dynamic state (accurately, including the change in mass) based on the known expenditure of propellant.  The fact that you can do neither, and don't even try, indicates you don't know what you're doing.  Your oversimplification is fatal to your claim.

Until you demonstrate even minimal competence, there is no point dissecting the rest of your "presentation" in depth.

as NASA cannot inform how much fuel or energy was consumed to reduce the speed from 2400 to 1500 m/s to get into orbit.

Factually false.  Detailed pads for all burns have been available for more than 10 years.  Someone so inept in his research does not enjoy a priori credibility.  You bear the burden to prove you have solved the problem correctly, and you frankly admit you have not.  It is not the job of your critics to educate you properly in the correct principles of spacecraft dynamics.  You have claimed to be an expert.  You will either therefore demonstrate expertise to our satisfaction, or you will be dismissed.

FGS, just read my presentation where all info is given.

Your presentation is undocumented and proceeds from false premises and pretenses that I have outlined and asked you to correct here.  Until those are corrected here, you have no reason to compel others to read a lengthy page of nonsense.  Your egregious mistakes are made early enough on that the rest of your "presentation" is nonsense.

[LunarOrbit]

I think Grashtel deserves the 1 million Euro.

Split 50/50 with Jay.

[Heiwa]

The above is basic - now try to show the errors in my presentation.

You used the wrong equations and made-up values for the quantities expressed by the equations you did use.  No further discussion is possible until you correct those errors.  In fact, when one uses the wrong model and the wrong initial values, there is not much more to the problem to get wrong.

No, I use the correct, but simple, equations and values obtained from NASA reports to get a feel of the problem as shown in my presentation. Pls show your equations and values to obtain the energy/force/time, etc, required to get into Moon orbit on arrival Moon and out of Moon orbit on departure Moon, so we can discuss seriously. 

[Heiwa]

Your presentation is undocumented and proceeds from false premises and pretenses that I have outlined and asked you to correct here.  Until those are corrected here, you have no reason to compel others to read a lengthy page of nonsense.  Your egregious mistakes are made early enough on that the rest of your "presentation" is nonsense.

No, my presentation is documented as references are given at start of presentation and values used are taken from these references and all calculations are correct. Of course the NASA references are full of errors some of which I point out. Nobody is compelled to read my presentation or to get upset about it.

If you want to win €1 000 000:- (topic) you just have to do your own calculations of energy (fuel) required and present them, e.g. copy/paste from a suitable NASA report. Shouldn't that be easy?

[JayUtah]

No, I use the correct, but simple, equations...

No.  You simplified away the most important part of the relevant equations -- the terms that deal with the change in spacecraft mass as fuel is consumed.  Use the correct equations.

...and values obtained from NASA reports...

As previously stated, you give velocity figures without citation ("according to NASA" is not sufficient documentation), and without placing them in an appropriate orbital dynamics context.  As such they are useless in computation.  I have asked you to correct those errors.  Please explain why you have not done so, and correct them immediately.

Pls show your equations...

There are no "my" equations; there are only the proper equations.  Your "simplifications" based on your obviously limited knowledge of astrodynamics do not suffice.  I've given you the hint:  you must consider the natural logarithm of the ratio of start and end masses.

I will not spoon feed you information that you should know according to your claim to be an engineer.  You have the burden to show you know what you're talking about.  I will give you a reference however to Sutton and Biblarz as authors of some note on the subject.  Reconcile your claims with the first few chapters in any of their books and return here.

Also, I will require proof that you are prepared to pay one million Euros on demand or else this conversation will not proceed.  Please deposit the sum in escrow in a bank of your choice and post its account number here along with the name of the escrow agent, and (in a private message) the PIN to verify the amount.

[JayUtah]

No, my presentation is documented as references are given at start of presentation...

No.  I have asked you specific questions about the context of your quoted values.  You are unable to answer them, indicating you do not understand them.

...all calculations are correct.

No.  By your own admission the spacecraft is a variable-mass vehicle, and you have specifically omitted that property in your computations.  You have already admitted that you are using a simplified model, not a correct model.  Either use the model you know you to be correct, or perform an error analysis to show that the difference between the correct model and your simplification is insignificant.

Nobody is compelled to read my presentation or to get upset about it.

Nobody does read it.  Claims that Apollo missions are phony, especially those based on admittedly imprecise computations, are simply dismissed as absurd in the industry.  Since you have challenged an entire industry and its subordinate sciences based upon nothing but your personal say-so, you bear considerable responsibility to answer questions and defend the basis of your claims.  Trying to shift the burden of proof to force your critics to educate you is profoundly unfair.  You are hubristically claiming superior understanding.  You will therefore demonstrate it at my request or else concede.

I have pointed out the initial errors in your presentation.  The rest of it is pointless verbiage until you correct those basic errors.

If you want to win €1 000 000:- (topic) you just have to do your own calculations of energy (fuel) required and present them, e.g. copy/paste from a suitable NASA report. Shouldn't that be easy?

No.  Your offer is to show what you did wrong.  We have done that.  You are obviously unwilling and unable to pay up.

[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.

[Heiwa]

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?

[Heiwa]

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 and get into orbit there in order not to crash on or simply fly bye, and according NASA you slow down from 2400 to 1500 m/s and for that you need 75.47 GJ energy (assuming constant mass 43 000 kg while slowing down).
As it seems 1 kg rocket fuel produces 1.63 MJ energy you need 46 300 kg fuel to slow down. The question is, where to store it?

The fuel burnt in the Service Module rocket engine evidently produces a force, unit Newton, that slows down the space ship from 2400 to 1500 m/s during a certain time t (seconds), while the space ship moves a certain trajectory/distance (meter). During this deceleration maneuver also the direction of the space ship is changed probably also helped by Moon gravity. Another question is how to control the direction of the force during this maneuver so that you neither crash nor fly bye the Moon. Any ideas?

[Andromeda]

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 and get into orbit there in order not to crash on or simply fly bye, and according NASA you slow down from 2400 to 1500 m/s and for that you need 75.47 GJ energy (assuming constant mass 43 000 kg while slowing down).
As it seems 1 kg rocket fuel produces 1.63 MJ energy you need 46 300 kg fuel to slow down. The question is, very to store it?

Two things.

1. Assuming constant mass is incorrect, as Jay already pointed out.  Jay is an aerospace engineer.

2. Even if using 1/2 m v² was right (it's not) you have done the operations in the wrong order to get your answer.  You have calculated delta(v²) when it should be (delta v)² as Chew already explained.

If you refuse to accept corrections to your mistakes from experts in the field, there is no point debating you.  I'd be annoyed if I wasn't laughing so hard.  A child would not make such mistakes as you have.

Further, I do not believe you have a million Euros, or would be willing to pay it over if you did.

Lastly, it is considered very poor form here to start posting new questions and demands while ignoring old ones.

[nomuse]

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?

1)  This was already answered in this thread.

2) This isn't some secret flaw NASA has been careful not to mention.  It is a well-documented part of the spacecraft operations.  Described in detail by Walter Cronkite to the world audience and all.

[Heiwa]

Nobody does read it.  Claims that Apollo missions are phony, especially those based on admittedly imprecise computations, are simply dismissed as absurd in the industry.  Since you have challenged an entire industry and its subordinate sciences based upon nothing but your personal say-so, you bear considerable responsibility to answer questions and defend the basis of your claims.  Trying to shift the burden of proof to force your critics to educate you is profoundly unfair.  You are hubristically claiming superior understanding.  You will therefore demonstrate it at my request or else concede.

I have pointed out the initial errors in your presentation.  The rest of it is pointless verbiage until you correct those basic errors.

If you want to win €1 000 000:- (topic) you just have to do your own calculations of energy (fuel) required and present them, e.g. copy/paste from a suitable NASA report. Shouldn't that be easy?

No.  Your offer is to show what you did wrong.  We have done that.  You are obviously unwilling and unable to pay up.

Actually Chew read it and ask real questions.

In order to win €1 million (topic) you must evidently use the correct equations, masses, velocities, forces, trajectories, times, etc to explain the Moon trip. As NASA has done it several times it seems you only have to copy/paste from the NASA reports and there  you are. I have not been able to do it, maybe because I am a stupid engineer? Or a conspiracy terrorist?
Actually I am neither. I am just curious how much energy you need to do do the trip and where to store it.

[Heiwa]

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 and get into orbit there in order not to crash on or simply fly bye, and according NASA you slow down from 2400 to 1500 m/s and for that you need 75.47 GJ energy (assuming constant mass 43 000 kg while slowing down).
As it seems 1 kg rocket fuel produces 1.63 MJ energy you need 46 300 kg fuel to slow down. The question is, very to store it?

Two things.

1. Assuming constant mass is incorrect, as Jay already pointed out.  Jay is an aerospace engineer.

2. Even if using 1/2 m v² was right (it's not) you have done the operations in the wrong order to get your answer.  You have calculated delta(v²) when it should be (delta v)² as Chew already explained.

If you refuse to accept corrections to your mistakes from experts in the field, there is no point debating you.  I'd be annoyed if I wasn't laughing so hard.  A child would not make such mistakes as you have.

Further, I do not believe you have a million Euros, or would be willing to pay it over if you did.

Lastly, it is considered very poor form here to start posting new questions and demands while ignoring old ones.

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.

[Heiwa]

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?

1)  This was already answered in this thread.

2) This isn't some secret flaw NASA has been careful not to mention.  It is a well-documented part of the spacecraft operations.  Described in detail by Walter Cronkite to the world audience and all.

?? 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?

[Andromeda]

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 and get into orbit there in order not to crash on or simply fly bye, and according NASA you slow down from 2400 to 1500 m/s and for that you need 75.47 GJ energy (assuming constant mass 43 000 kg while slowing down).
As it seems 1 kg rocket fuel produces 1.63 MJ energy you need 46 300 kg fuel to slow down. The question is, very to store it?

Two things.

1. Assuming constant mass is incorrect, as Jay already pointed out.  Jay is an aerospace engineer.

2. Even if using 1/2 m v² was right (it's not) you have done the operations in the wrong order to get your answer.  You have calculated delta(v²) when it should be (delta v)² as Chew already explained.

If you refuse to accept corrections to your mistakes from experts in the field, there is no point debating you.  I'd be annoyed if I wasn't laughing so hard.  A child would not make such mistakes as you have.

Further, I do not believe you have a million Euros, or would be willing to pay it over if you did.

Lastly, it is considered very poor form here to start posting new questions and demands while ignoring old ones.

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

You really don't see the difference between

2400^² - 1500^² = 3,510,000

And

(2400-1500)^2 = 810,000

Because that is but one error you have made.



Or that the mass changes as fuel is used?

??  Really?