"Thomas Heger" wrote in message
...
Am 18.10.2011 10:14, schrieb Scout:


"Thomas Heger" wrote in message
...
Am 17.10.2011 07:01, schrieb Scout:


..

For Earth orbit you need much faster flight than you would need to
stay in orbit around the moon, but nevertheless it is quite fast. On
Earth it took a Saturn V rocket, to lift the craft into orbit. On the
Moon it would take less fuel, but way more, than the few gallons, they
had in the lander.

Ok, let's see your math.

I mean if you know they needed more, then clearly you have calculated
all this out and know exactly how much they would need and whether they
could have that much on the lander.

So let's see your work.

---- Insert mathematical proof here.

Here I will even aid you with the specifications for the mass, amount
of
fuel, type of fuel, specific impulse, thrust provided, available
delta-V, and so on.

http://en.wikipedia.org/wiki/Apollo_...Specifications


Well, I'm a little too lazy, but a rough calculation is possible:

There is the Tsiolkovsky rocket equation
velocity_final=v_exhaust* ln(mass_start/mass_finish)

V_end= 2200 m/s * ln (4547 kg/(4547-2353) kg)

that is :
v_end approx. 1603 m/s

this is an estimated calculation without gravity.

the final velocity is reduced by
delta v = g_moon * (time of engine running)

Don't know that number (time_ engine)

Maybe 100 seconds (???)

makes:
delta v = 1.6 m/s²*100 s=160 m/s

What gives a rough estimate for the final velocity of the landers
ascending stage of
v_end = 1440 m/s.

Now the orbital velocity had to be compared. But I don't have the data
and actually I'm too lazy to find them out. But usual orbits should be
a little less than escape velocity, what is
v_orbit_escape = 2380 m/s.

V_end is a rough estimate ('thumb times pi'). For better calculations
someone with more experience in rocket science is needed.

I cannot even tell you, if the ascent stage is fast enough or not. But
my intuition tells me, it is not.


IOW, you don't know what the hell you're talking about, and you're too
lazy to do the work needed to find out if what you think actually has
merit or is simply bat **** crazy.


I haven't claimed to be a rocket scientist. I'm totally happy with an
rough estimate. I could do it better, for sure, but do not want.

What you have isn't even a rought estimate that applies. You simply threw
some stuff up there, came up with some answers, but didn't use the data from
the apollo program, which it should be noted I was even nice enough to lead
you to by the hand, much less show that the results produced proved that a
landing and take-off physically could not occur given those conditions. You
simply flopped around trying to put together an argument.

Free hint: If you're going to say someone else is lying, then you need to
make sure you have your ducks in a row and can PROVE IT.

All you've shown is that you are an empty headed conspiracy theorist, with
lots of notions, but no facts, no proof, and from all evidence absolutely NO
desire to find out what the facts really are.

snip

Am 18.10.2011 19:42, schrieb Scout:


"Thomas Heger" wrote in message
...
Am 18.10.2011 10:14, schrieb Scout:


"Thomas Heger" wrote in message
...
Am 17.10.2011 07:01, schrieb Scout:


..

For Earth orbit you need much faster flight than you would need to
stay in orbit around the moon, but nevertheless it is quite fast. On
Earth it took a Saturn V rocket, to lift the craft into orbit. On the
Moon it would take less fuel, but way more, than the few gallons,
they
had in the lander.

Ok, let's see your math.

I mean if you know they needed more, then clearly you have calculated
all this out and know exactly how much they would need and whether
they
could have that much on the lander.

So let's see your work.

---- Insert mathematical proof here.

Here I will even aid you with the specifications for the mass,
amount of
fuel, type of fuel, specific impulse, thrust provided, available
delta-V, and so on.

http://en.wikipedia.org/wiki/Apollo_...Specifications


Well, I'm a little too lazy, but a rough calculation is possible:

There is the Tsiolkovsky rocket equation
velocity_final=v_exhaust* ln(mass_start/mass_finish)

V_end= 2200 m/s * ln (4547 kg/(4547-2353) kg)

that is :
v_end approx. 1603 m/s

this is an estimated calculation without gravity.

the final velocity is reduced by
delta v = g_moon * (time of engine running)

Don't know that number (time_ engine)

Maybe 100 seconds (???)

makes:
delta v = 1.6 m/s²*100 s=160 m/s

What gives a rough estimate for the final velocity of the landers
ascending stage of
v_end = 1440 m/s.

Now the orbital velocity had to be compared. But I don't have the data
and actually I'm too lazy to find them out. But usual orbits should be
a little less than escape velocity, what is
v_orbit_escape = 2380 m/s.

V_end is a rough estimate ('thumb times pi'). For better calculations
someone with more experience in rocket science is needed.

I cannot even tell you, if the ascent stage is fast enough or not. But
my intuition tells me, it is not.


IOW, you don't know what the hell you're talking about, and you're too
lazy to do the work needed to find out if what you think actually has
merit or is simply bat **** crazy.


I haven't claimed to be a rocket scientist. I'm totally happy with an
rough estimate. I could do it better, for sure, but do not want.

What you have isn't even a rought estimate that applies. You simply
threw some stuff up there, came up with some answers, but didn't use the
data from the apollo program, which it should be noted I was even nice
enough to lead you to by the hand, much less show that the results
produced proved that a landing and take-off physically could not occur
given those conditions. You simply flopped around trying to put together
an argument.

Free hint: If you're going to say someone else is lying, then you need
to make sure you have your ducks in a row and can PROVE IT.

All you've shown is that you are an empty headed conspiracy theorist,
with lots of notions, but no facts, no proof, and from all evidence
absolutely NO desire to find out what the facts really are.

You are absolutely wrong!
The rocket equation is a method to calculate the final velocity of a
single staged rocket.

The ascent stage would fit to 'rocket', even if doesn't look like. It
had - of course - only one stage.

The rocket equation ignores gravity. The moon has low gravity, what
makes this equation even more usable.

The precise orbit of the orbiter I could figure out, but that would be
'work', while typing stuff into the UseNet qualifies as 'leisure'.

So I decided, I don't want to do that. This decision is absolutely my
right and nobody could hold me responsible, because I refuse to
calculate the orbit of the command module of Apollo 11.


TH

On 10/18/2011 11:56 AM, Thomas Heger wrote:
Am 18.10.2011 19:42, schrieb Scout:


"Thomas Heger" wrote in message
...
Am 18.10.2011 10:14, schrieb Scout:


"Thomas Heger" wrote in message
...
Am 17.10.2011 07:01, schrieb Scout:


..

For Earth orbit you need much faster flight than you would need to
stay in orbit around the moon, but nevertheless it is quite fast. On
Earth it took a Saturn V rocket, to lift the craft into orbit. On
the
Moon it would take less fuel, but way more, than the few gallons,
they
had in the lander.

Ok, let's see your math.

I mean if you know they needed more, then clearly you have calculated
all this out and know exactly how much they would need and whether
they
could have that much on the lander.

So let's see your work.

---- Insert mathematical proof here.

Here I will even aid you with the specifications for the mass,
amount of
fuel, type of fuel, specific impulse, thrust provided, available
delta-V, and so on.

http://en.wikipedia.org/wiki/Apollo_...Specifications


Well, I'm a little too lazy, but a rough calculation is possible:

There is the Tsiolkovsky rocket equation
velocity_final=v_exhaust* ln(mass_start/mass_finish)

V_end= 2200 m/s * ln (4547 kg/(4547-2353) kg)

that is :
v_end approx. 1603 m/s

this is an estimated calculation without gravity.

the final velocity is reduced by
delta v = g_moon * (time of engine running)

Don't know that number (time_ engine)

Maybe 100 seconds (???)

makes:
delta v = 1.6 m/s²*100 s=160 m/s

What gives a rough estimate for the final velocity of the landers
ascending stage of
v_end = 1440 m/s.

Now the orbital velocity had to be compared. But I don't have the data
and actually I'm too lazy to find them out. But usual orbits should be
a little less than escape velocity, what is
v_orbit_escape = 2380 m/s.

V_end is a rough estimate ('thumb times pi'). For better calculations
someone with more experience in rocket science is needed.

I cannot even tell you, if the ascent stage is fast enough or not. But
my intuition tells me, it is not.


IOW, you don't know what the hell you're talking about, and you're too
lazy to do the work needed to find out if what you think actually has
merit or is simply bat **** crazy.


I haven't claimed to be a rocket scientist. I'm totally happy with an
rough estimate. I could do it better, for sure, but do not want.

What you have isn't even a rought estimate that applies. You simply
threw some stuff up there, came up with some answers, but didn't use the
data from the apollo program, which it should be noted I was even nice
enough to lead you to by the hand, much less show that the results
produced proved that a landing and take-off physically could not occur
given those conditions. You simply flopped around trying to put together
an argument.

Free hint: If you're going to say someone else is lying, then you need
to make sure you have your ducks in a row and can PROVE IT.

All you've shown is that you are an empty headed conspiracy theorist,
with lots of notions, but no facts, no proof, and from all evidence
absolutely NO desire to find out what the facts really are.

You are absolutely wrong!
The rocket equation is a method to calculate the final velocity of a
single staged rocket.

The ascent stage would fit to 'rocket', even if doesn't look like. It
had - of course - only one stage.

The rocket equation ignores gravity. The moon has low gravity, what
makes this equation even more usable.

The precise orbit of the orbiter I could figure out, but that would be
'work', while typing stuff into the UseNet qualifies as 'leisure'.

So I decided, I don't want to do that. This decision is absolutely my
right and nobody could hold me responsible, because I refuse to
calculate the orbit of the command module of Apollo 11.


TH



You need to review the past posts of "the scout", this is just a mental
case attempting to be a troll ... hang it up ... have discussions with
lunatics is never fruitful ... although they might be fruits ...

Regards,
JS

"Thomas Heger" wrote in message
...
Am 18.10.2011 19:42, schrieb Scout:


"Thomas Heger" wrote in message
...
Am 18.10.2011 10:14, schrieb Scout:


"Thomas Heger" wrote in message
...
Am 17.10.2011 07:01, schrieb Scout:


..

For Earth orbit you need much faster flight than you would need to
stay in orbit around the moon, but nevertheless it is quite fast. On
Earth it took a Saturn V rocket, to lift the craft into orbit. On
the
Moon it would take less fuel, but way more, than the few gallons,
they
had in the lander.

Ok, let's see your math.

I mean if you know they needed more, then clearly you have calculated
all this out and know exactly how much they would need and whether
they
could have that much on the lander.

So let's see your work.

---- Insert mathematical proof here.

Here I will even aid you with the specifications for the mass,
amount of
fuel, type of fuel, specific impulse, thrust provided, available
delta-V, and so on.

http://en.wikipedia.org/wiki/Apollo_...Specifications


Well, I'm a little too lazy, but a rough calculation is possible:

There is the Tsiolkovsky rocket equation
velocity_final=v_exhaust* ln(mass_start/mass_finish)

V_end= 2200 m/s * ln (4547 kg/(4547-2353) kg)

that is :
v_end approx. 1603 m/s

this is an estimated calculation without gravity.

the final velocity is reduced by
delta v = g_moon * (time of engine running)

Don't know that number (time_ engine)

Maybe 100 seconds (???)

makes:
delta v = 1.6 m/s²*100 s=160 m/s

What gives a rough estimate for the final velocity of the landers
ascending stage of
v_end = 1440 m/s.

Now the orbital velocity had to be compared. But I don't have the data
and actually I'm too lazy to find them out. But usual orbits should be
a little less than escape velocity, what is
v_orbit_escape = 2380 m/s.

V_end is a rough estimate ('thumb times pi'). For better calculations
someone with more experience in rocket science is needed.

I cannot even tell you, if the ascent stage is fast enough or not. But
my intuition tells me, it is not.


IOW, you don't know what the hell you're talking about, and you're too
lazy to do the work needed to find out if what you think actually has
merit or is simply bat **** crazy.


I haven't claimed to be a rocket scientist. I'm totally happy with an
rough estimate. I could do it better, for sure, but do not want.

What you have isn't even a rought estimate that applies. You simply
threw some stuff up there, came up with some answers, but didn't use the
data from the apollo program, which it should be noted I was even nice
enough to lead you to by the hand, much less show that the results
produced proved that a landing and take-off physically could not occur
given those conditions. You simply flopped around trying to put together
an argument.

Free hint: If you're going to say someone else is lying, then you need
to make sure you have your ducks in a row and can PROVE IT.

All you've shown is that you are an empty headed conspiracy theorist,
with lots of notions, but no facts, no proof, and from all evidence
absolutely NO desire to find out what the facts really are.

You are absolutely wrong!

Then let's see your proof. Not some bull**** hack job you threw together,
but conclusive factual objective mathematical PROOF.....

Otherwise, I'm right.


The rocket equation is a method to calculate the final velocity of a
single staged rocket.

The ascent stage would fit to 'rocket', even if doesn't look like. It
had - of course - only one stage.

The rocket equation ignores gravity. The moon has low gravity, what makes
this equation even more usable.

The precise orbit of the orbiter I could figure out, but that would be
'work', while typing stuff into the UseNet qualifies as 'leisure'.

See what I mean.....no desire to find/figure out the actual facts.

So I decided, I don't want to do that.

IOW, screw figuring out the facts.

This decision is absolutely my right

Absolutely, you have every right to be as stupid and ignorant as you
chose....but don't confuse that with being informed.


and nobody could hold me responsible, because I refuse to calculate the
orbit of the command module of Apollo 11.

So basically, you've got a lot of hot air....but nothing to support it.

Typical conspiracy theorist....long on talk, short on facts.

Am 19.10.2011 10:42, schrieb Scout:



IOW, you don't know what the hell you're talking about, and you're too
lazy to do the work needed to find out if what you think actually has
merit or is simply bat **** crazy.


I haven't claimed to be a rocket scientist. I'm totally happy with an
rough estimate. I could do it better, for sure, but do not want.

What you have isn't even a rought estimate that applies. You simply
threw some stuff up there, came up with some answers, but didn't use the
data from the apollo program, which it should be noted I was even nice
enough to lead you to by the hand, much less show that the results
produced proved that a landing and take-off physically could not occur
given those conditions. You simply flopped around trying to put together
an argument.

Free hint: If you're going to say someone else is lying, then you need
to make sure you have your ducks in a row and can PROVE IT.

All you've shown is that you are an empty headed conspiracy theorist,
with lots of notions, but no facts, no proof, and from all evidence
absolutely NO desire to find out what the facts really are.

You are absolutely wrong!

Then let's see your proof. Not some bull**** hack job you threw
together, but conclusive factual objective mathematical PROOF.....

Otherwise, I'm right.

OK. You seem to insist on something more profound ;-(.

As I said, that needs some sort of work. So I relocate from sofa to
desktop and try to figure that out. But - please - a few days are
necessary, what is certainly possible ?

TH

Sofa to desktop? Doggy's couch (sofa) is my WebTV comfort zone.
cuhulin

http://www.devilfinder.com/find.php?...es+Saved+D-Day
See that hand cranked Tides Calculator Machine?

Some American G.Is drowned when they landed, stepped into the water,
because they were carrying so much weight in their backpacks.

The old Higgins Boat factory, part of the D-Day Museum in New Orleans
sits on the site of that old Boat Factory.
cuhulin

On 10/18/2011 10:42 AM, Scout wrote:


"Thomas Heger" wrote in message
...
Am 18.10.2011 10:14, schrieb Scout:


"Thomas Heger" wrote in message
...
Am 17.10.2011 07:01, schrieb Scout:


..

For Earth orbit you need much faster flight than you would need to
stay in orbit around the moon, but nevertheless it is quite fast. On
Earth it took a Saturn V rocket, to lift the craft into orbit. On the
Moon it would take less fuel, but way more, than the few gallons,
they
had in the lander.

Ok, let's see your math.

I mean if you know they needed more, then clearly you have calculated
all this out and know exactly how much they would need and whether
they
could have that much on the lander.

So let's see your work.

---- Insert mathematical proof here.

Here I will even aid you with the specifications for the mass,
amount of
fuel, type of fuel, specific impulse, thrust provided, available
delta-V, and so on.

http://en.wikipedia.org/wiki/Apollo_...Specifications


Well, I'm a little too lazy, but a rough calculation is possible:

There is the Tsiolkovsky rocket equation
velocity_final=v_exhaust* ln(mass_start/mass_finish)

V_end= 2200 m/s * ln (4547 kg/(4547-2353) kg)

that is :
v_end approx. 1603 m/s

this is an estimated calculation without gravity.

the final velocity is reduced by
delta v = g_moon * (time of engine running)

Don't know that number (time_ engine)

Maybe 100 seconds (???)

makes:
delta v = 1.6 m/s²*100 s=160 m/s

What gives a rough estimate for the final velocity of the landers
ascending stage of
v_end = 1440 m/s.

Now the orbital velocity had to be compared. But I don't have the data
and actually I'm too lazy to find them out. But usual orbits should be
a little less than escape velocity, what is
v_orbit_escape = 2380 m/s.

V_end is a rough estimate ('thumb times pi'). For better calculations
someone with more experience in rocket science is needed.

I cannot even tell you, if the ascent stage is fast enough or not. But
my intuition tells me, it is not.


IOW, you don't know what the hell you're talking about, and you're too
lazy to do the work needed to find out if what you think actually has
merit or is simply bat **** crazy.


I haven't claimed to be a rocket scientist. I'm totally happy with an
rough estimate. I could do it better, for sure, but do not want.

What you have isn't even a rought estimate that applies. You simply
threw some stuff up there, came up with some answers, but didn't use the
data from the apollo program, which it should be noted I was even nice
enough to lead you to by the hand, much less show that the results
produced proved that a landing and take-off physically could not occur
given those conditions. You simply flopped around trying to put together
an argument.

Free hint: If you're going to say someone else is lying, then you need
to make sure you have your ducks in a row and can PROVE IT.

All you've shown is that you are an empty headed conspiracy theorist,
with lots of notions, but no facts, no proof, and from all evidence
absolutely NO desire to find out what the facts really are.

snip



Since I met you, you were a loon, nothing has gotten better with time ...

Regards,
JS

"John Smith" wrote in message
...
On 10/18/2011 10:42 AM, Scout wrote:


"Thomas Heger" wrote in message
...
Am 18.10.2011 10:14, schrieb Scout:


"Thomas Heger" wrote in message
...
Am 17.10.2011 07:01, schrieb Scout:


..

For Earth orbit you need much faster flight than you would need to
stay in orbit around the moon, but nevertheless it is quite fast. On
Earth it took a Saturn V rocket, to lift the craft into orbit. On
the
Moon it would take less fuel, but way more, than the few gallons,
they
had in the lander.

Ok, let's see your math.

I mean if you know they needed more, then clearly you have calculated
all this out and know exactly how much they would need and whether
they
could have that much on the lander.

So let's see your work.

---- Insert mathematical proof here.

Here I will even aid you with the specifications for the mass,
amount of
fuel, type of fuel, specific impulse, thrust provided, available
delta-V, and so on.

http://en.wikipedia.org/wiki/Apollo_...Specifications


Well, I'm a little too lazy, but a rough calculation is possible:

There is the Tsiolkovsky rocket equation
velocity_final=v_exhaust* ln(mass_start/mass_finish)

V_end= 2200 m/s * ln (4547 kg/(4547-2353) kg)

that is :
v_end approx. 1603 m/s

this is an estimated calculation without gravity.

the final velocity is reduced by
delta v = g_moon * (time of engine running)

Don't know that number (time_ engine)

Maybe 100 seconds (???)

makes:
delta v = 1.6 m/s²*100 s=160 m/s

What gives a rough estimate for the final velocity of the landers
ascending stage of
v_end = 1440 m/s.

Now the orbital velocity had to be compared. But I don't have the data
and actually I'm too lazy to find them out. But usual orbits should be
a little less than escape velocity, what is
v_orbit_escape = 2380 m/s.

V_end is a rough estimate ('thumb times pi'). For better calculations
someone with more experience in rocket science is needed.

I cannot even tell you, if the ascent stage is fast enough or not. But
my intuition tells me, it is not.


IOW, you don't know what the hell you're talking about, and you're too
lazy to do the work needed to find out if what you think actually has
merit or is simply bat **** crazy.


I haven't claimed to be a rocket scientist. I'm totally happy with an
rough estimate. I could do it better, for sure, but do not want.

What you have isn't even a rought estimate that applies. You simply
threw some stuff up there, came up with some answers, but didn't use the
data from the apollo program, which it should be noted I was even nice
enough to lead you to by the hand, much less show that the results
produced proved that a landing and take-off physically could not occur
given those conditions. You simply flopped around trying to put together
an argument.

Free hint: If you're going to say someone else is lying, then you need
to make sure you have your ducks in a row and can PROVE IT.

All you've shown is that you are an empty headed conspiracy theorist,
with lots of notions, but no facts, no proof, and from all evidence
absolutely NO desire to find out what the facts really are.

snip



Since I met you, you were a loon, nothing has gotten better with time ...

Nope, I simply don't buy his bull**** without proof, just as I wouldn't buy
yours.

If you throw something out there, then you need to be able to back it up
with some support if called on it.

If you can't then expect to be treated in the manner your deserve.

Hand cranked Calculator, you said?

http://www.devilfinder.com/find.php?...ide+Calculator

http://www.ddaymuseum.org

When World War Two was over, America sent some big machines to
Germany/Europe.The machines were used for grinding up that rubble, the
ground up rubble was mixed into new concrete/rebuilding.

I have a Curta hand crank Calculator.I bought it for $88.00 at an Indian
(India) store in Saigon,Vietnam in 1964.
cuhulin