Merged Apollo "hoax" discussion / Lick observatory laser saga

[JayUtah]

On page 125 of David Whitehouse's book, ONE SMALL STEP(Quercus Publishing, London, 2009), Armstrong is quoted as saying that his concern about running out of fuel during his "lunar module landing" on 07/20/1969 was somewhat mitigated by his knowing that if he could get his speed and attitude stabilized, he could fall from a fairly good height "PERHAPS MAYBE 40 FEET OR MORE IN THE LOW LUNAR GRAVITY". Armstrong claimed the gear would absorb that much fall. As such, Armstrong claimed he wasn't as concerned as those back on earth, not as concerned about the gas running out problem.

Does Armstrong's claim make any sense? Well of course not. I'll show you why and then I will put the gaffe in a broader perspective.

The lunar lander's mass was 15,200 kg without the propellant. So it's earth weight was 33,440 lbs. From 40 feet up at 1/6 of the earth's gravity it would take 3.87 seconds for the Eagle to hit the surface of the moon were it to free fall. The kinetic energy the Eagle would possess were it to fall from 40 feet is given by the simple equation mass x gravity x height and in this case that would be 15,200kg X 9.81 m/s per s/6 x 12.192 meters = 302,996 joules.

So if we take Armstrong at his word, a 40 foot Eagle fall under lunar conditions would mean the 15,200 kg bird would strike the lunar surface after a 3.87 second fall at a velocity of 6.33 meters per second or equivalently 20.6 feet per second or equivalently 13.6 miles and hour. The Eagle's kinetic energy on impact would be 302,996 joules.

I'll translate that into more familiar terms. The average compact car weighs roughly 3000-4,500 lbs. I'll use the halfway point, 3,750 lbs(1704 kg) The Eagle weighs 33,440 lbs without "gas" so that's 8.92 times as much as the compact car. I'll show an equivalent earth based situation using the compact car as the falling object, equivalent in that at the time the compact hits the earth it will be carrying 302,996 joules of energy.

Here on earth the kinetic energy of a falling object will of course also be given by mass x gravity x height. One wants here the height of a compact car fall that will give a solution providing 302,996 joules of energy. 302,996 divided by 1704 divided by 9.8 = 18.14 meters or 59 feet. So if a compact car were to be dropped from 59 feet, it would take 3.48 seconds to hit the earth with the same kinetic energy as Neil Armstrong's Eagle falling from 40 feet on the moon. My compact car would be moving at 34 meters per second at the time of impact. That translates to 112 feet per second or 76 miles an hour.

So let's think about Armstrong's claim. He says that if he were to be able to control the Eagle's speed and attitude so that it simply dropped straight down, the thing could land on its legs and everything would be hunky dory.

Doesn't make any sense at all does it now? Put those lunar module legs on a compact car. Secure them any way you like. Run the car into a wall at 76 miles per hour. And, given Armstrong's scenario, what's the likelihood under those circumstances that the thing will hit with one and not 4 legs. A 3750 lb car moving at 76 miles an hour running an Apollo lunar module leg into a solid piece of rock.

So Armstrong is way wrong here, either that or the dude is stranded on the moon with Aldrin.

What's going on? Well it is just like when Collins says at the Apollo 11 post flight press conference that he couldn't recall seeing any stars when they traveled to the moon as the moon was eclipsing the sun. Same thing here. It is a ridiculous statement by Armstrong. He rarely messed up, but this here indeed was one of those rare screw ups. This of course is why Armstrong did not give interviews as a rule, did not write books, did not answer questions. The more of that stuff you do, the more this kind of stuff happens.

40 foot drop of a lunar module could easily break the leg on a lunar module and strand you with Aldrin.

Why are you changing the subject yet again? Aren't there enough open questions begging for your attention, Patrick? Are you so quickly backed into the corner again?

You just got done telling us in another thread that you weren't really a doctor as you previously claimed, and that it was all an elaborate ruse and we all fell for it and you were so clever and we should have been able to see through it because it was such an obvious deception. That pretty much destroys your claim to superior expertise in the Borman illness claim, which you said was your slam-dunk.

Will you please address the withdrawal of your claim to expertise?

 

[TjW]

On page 125 of David Whitehouse's book, ONE SMALL STEP(Quercus Publishing, London, 2009), Armstrong is quoted as saying that his concern about running out of fuel during his "lunar module landing" on 07/20/1969 was somewhat mitigated by his knowing that if he could get his speed and attitude stabilized, he could fall from a fairly good height "PERHAPS MAYBE 40 FEET OR MORE IN THE LOW LUNAR GRAVITY". Armstrong claimed the gear would absorb that much fall. As such, Armstrong claimed he wasn't as concerned as those back on earth, not as concerned about the gas running out problem.

Does Armstrong's claim make any sense? Well of course not. I'll show you why and then I will put the gaffe in a broader perspective.

The lunar lander's mass was 15,200 kg without the propellant. So it's earth weight was 33,440 lbs. From 40 feet up at 1/6 of the earth's gravity it would take 3.87 seconds for the Eagle to hit the surface of the moon were it to free fall. The kinetic energy the Eagle would possess were it to fall from 40 feet is given by the simple equation mass x gravity x height and in this case that would be 15,200kg X 9.81 m/s per s/6 x 12.192 meters = 302,996 joules.

So if we take Armstrong at his word, a 40 foot Eagle fall under lunar conditions would mean the 15,200 kg bird would strike the lunar surface after a 3.87 second fall at a velocity of 6.33 meters per second or equivalently 20.6 feet per second or equivalently 13.6 miles and hour. The Eagle's kinetic energy on impact would be 302,996 joules.

Bzzzt! Wrong. Thanks for playing.

 

[Patrick1000]

A Little Common Sense Proves Apollo Inauthenticity

I'm sure many here are more than familiar with the famous Charles Sonett line about NASA's manned space flight personal, the scientists with the program at its inception and the Apollo program administrators. Sonett had designed experiments for the Pioneer craft while working for Space Technology Labs. When things were not going all that well at JPL with the design/performance of the Ranger craft, NASA hired Sonett as their chief lunar and planetary scientist. Sonett was a practical man, very smart, in more ways than one. He said of the folks running the Apollo effort;

"They were real dizzy….they had no idea what to really do, because they had no sense of science. Not one of 'em knew anything about the Moon. I am not sure they knew how to spell 'moon' ".


And when one pauses to think about it for a moment, Sonett's comment makes more than a little sense. Recall how von Braun originally had "planned" to not employ a lunar orbital rendezvous technique. The alleged original lunar landing plan was to go from earth orbit and cross cislunar space, park the spacecraft on the moon, and then blast off and head back to earth. Of course the problem with that is one must carry all the gas needed to return to earth down to the moon, along with the tanks for carrying that gas, along with a bigger engine needed to land all that gas on the moon and lift it back off the lunar surface.

Do the math…….The rocket leaving the planet earth would have to be so insanely big to pull off a NON lunar orbital rendezvous landing. The whole thing is more than beyond ludicrous. One can hardly keep oneself from laughing out loud upon consideration of this ridiculous plan's having been seriously offered up as in any sense viable. "Dizzy"? you betcha'.... No wonder Sonett said it of his Manned Space Flight Center colleagues.

Despite Sonett's opinion of the Apollo boys, von Braun was no fool and he of course knew from the get go they could never build a space ship big enough to get a landing craft down to the moon's surface that would have to carry so much gas and such a big engine such that it would be capable of blasting off from the moon and traveling all the way back to earth. von Braun could not possibly have offered up such an idiotic original plan with a straight face. Well he had a straight face, but that is because most of those around him were indeed "dizzy" as colleague Sonett called the Apollo manned landing team.

This is why Houbolt may be viewed provisionally as a plant, a fraud insider. He was the engineer that proposed lunar orbital rendezvous as the method of choice in getting the "astronauts" down to the lunar surface. It is semi-rational unlike von Braun's dizzy first pitch, and of course it provides for the construction and utilization of a piece of military equipment, the lunar module, that would be needed in carrying out the program's unmanned nefarious objectives.

 

[Patrick1000]

I have brought this subject up more than once before.....

Why are you changing the subject yet again? Aren't there enough open questions begging for your attention, Patrick? Are you so quickly backed into the corner again?

You just got done telling us in another thread that you weren't really a doctor as you previously claimed, and that it was all an elaborate ruse and we all fell for it and you were so clever and we should have been able to see through it because it was such an obvious deception. That pretty much destroys your claim to superior expertise in the Borman illness claim, which you said was your slam-dunk.

Will you please address the withdrawal of your claim to expertise?

I have brought this subject up more than once before Jay, about Armstrong avoiding interviews and so forth to not get caught making such a gaffe as that referenced above.

I was crystal clear with respect to my comments in the other thread that it was SATIRE, it could not have been read any other way. Review it for yourself. The tale was Paul Bunyanesque and ridiculous at every turn, and I said so directly. I said it was SATIRE. I even posted a photo of Paul Bunyan!

Of course I am a doctor.

We should get back to debating the FACTS of Apollo Jay. And i am entitled to introduce any subject material relevant that I choose. My point about the Armstrong gaffe above is a most excellent one, as are all of my other points as well.

 

[fess]

On page 125 of David Whitehouse's book, ONE SMALL STEP(Quercus Publishing, London, 2009), Armstrong is quoted as saying that his concern about running out of fuel during his "lunar module landing" on 07/20/1969 was somewhat mitigated by his knowing that if he could get his speed and attitude stabilized, he could fall from a fairly good height "PERHAPS MAYBE 40 FEET OR MORE IN THE LOW LUNAR GRAVITY". Armstrong claimed the gear would absorb that much fall. As such, Armstrong claimed he wasn't as concerned as those back on earth, not as concerned about the gas running out problem.

Does Armstrong's claim make any sense? Well of course not. I'll show you why and then I will put the gaffe in a broader perspective.

Did they run out of fuel? Did they fall forty feet? If not, the rest of your post is meaningless.

 

[Border Reiver]

On page 125 of David Whitehouse's book, ONE SMALL STEP(Quercus Publishing, London, 2009), Armstrong is quoted as saying that his concern about running out of fuel during his "lunar module landing" on 07/20/1969 was somewhat mitigated by his knowing that if he could get his speed and attitude stabilized, he could fall from a fairly good height "PERHAPS MAYBE 40 FEET OR MORE IN THE LOW LUNAR GRAVITY". Armstrong claimed the gear would absorb that much fall. As such, Armstrong claimed he wasn't as concerned as those back on earth, not as concerned about the gas running out problem.

Does Armstrong's claim make any sense? Well of course not. I'll show you why and then I will put the gaffe in a broader perspective.

The lunar lander's mass was 15,200 kg without the propellant. So it's earth weight was 33,440 lbs. From 40 feet up at 1/6 of the earth's gravity it would take 3.87 seconds for the Eagle to hit the surface of the moon were it to free fall. The kinetic energy the Eagle would possess were it to fall from 40 feet is given by the simple equation mass x gravity x height and in this case that would be 15,200kg X 9.81 m/s per s/6 x 12.192 meters = 302,996 joules.

So if we take Armstrong at his word, a 40 foot Eagle fall under lunar conditions would mean the 15,200 kg bird would strike the lunar surface after a 3.87 second fall at a velocity of 6.33 meters per second or equivalently 20.6 feet per second or equivalently 13.6 miles and hour. The Eagle's kinetic energy on impact would be 302,996 joules.

I'll translate that into more familiar terms. The average compact car weighs roughly 3000-4,500 lbs. I'll use the halfway point, 3,750 lbs(1704 kg) The Eagle weighs 33,440 lbs without "gas" so that's 8.92 times as much as the compact car. I'll show an equivalent earth based situation using the compact car as the falling object, equivalent in that at the time the compact hits the earth it will be carrying 302,996 joules of energy.

Here on earth the kinetic energy of a falling object will of course also be given by mass x gravity x height. One wants here the height of a compact car fall that will give a solution providing 302,996 joules of energy. 302,996 divided by 1704 divided by 9.8 = 18.14 meters or 59 feet. So if a compact car were to be dropped from 59 feet, it would take 3.48 seconds to hit the earth with the same kinetic energy as Neil Armstrong's Eagle falling from 40 feet on the moon. My compact car would be moving at 34 meters per second at the time of impact. That translates to 112 feet per second or 76 miles an hour.

So let's think about Armstrong's claim. He says that if he were to be able to control the Eagle's speed and attitude so that it simply dropped straight down, the thing could land on its legs and everything would be hunky dory.

Doesn't make any sense at all does it now? Put those lunar module legs on a compact car. Secure them any way you like. Run the car into a wall at 76 miles per hour. And, given Armstrong's scenario, what's the likelihood under those circumstances that the thing will hit with one and not 4 legs. A 3750 lb car moving at 76 miles an hour running an Apollo lunar module leg into a solid piece of rock.

So Armstrong is way wrong here, either that or the dude is stranded on the moon with Aldrin.

What's going on? Well it is just like when Collins says at the Apollo 11 post flight press conference that he couldn't recall seeing any stars when they traveled to the moon as the moon was eclipsing the sun. Same thing here. It is a ridiculous statement by Armstrong. He rarely messed up, but this here indeed was one of those rare screw ups. This of course is why Armstrong did not give interviews as a rule, did not write books, did not answer questions. The more of that stuff you do, the more this kind of stuff happens.

40 foot drop of a lunar module could easily break the leg on a lunar module and strand you with Aldrin.

Go back and get a refund for your physics and maths degrees Pat - they taught you the wrong formula for deriving kinetic energy.

Kinetic energy is derived from the formula e=1/2m*v² where m is the mass in kg and v is the velocity in metres per second.

Using your velocity for the drop of the lunar lander, we get

e = 7600* 40.0689 = 304,523.64 joules of energy

Now your compact car, the speed of impact is a lot different from what you've derived:

In this case we're solving for v, so your formula is

v² = e/(1/2 m)

or

v = square root of the energy divided by one half the mass

using your mass for the car, I derive a velocity of 13.366 m/s or 29.83178 miles per hour.

All this of course is an exercise in remembering my high school physics from 1986 and pointless unless I knew what is the level of energy that the legs of the lander were designed to withstand?

 

[haibut]

Here on earth the kinetic energy of a falling object will of course also be given by mass x gravity x height. One wants here the height of a compact car fall that will give a solution providing 302,996 joules of energy. 302,996 divided by 1704 divided by 9.8 = 18.14 meters or 59 feet. So if a compact car were to be dropped from 59 feet, it would take 3.48

Why a compact car and not a boeing 747? What happens if we use a 747 with a maximum takeoff weight of 333,390 Kg?

Here on earth the kinetic energy of a falling object will of course also be given by mass x gravity x height. One wants here the height of a Boeing 747 fall that will give a solution providing 302,996 joules of energy. 302,996 divided by 333,390 divided by 9.8 = 0.09 meters or 3.5 inches.

So you can't land a 747 from 3 1/2 inches.

 

[abaddon]

The lunar lander's mass was 15,200 kg without the propellant. So it's earth weight was 33,440 lbs. From 40 feet up at 1/6 of the earth's gravity it would take 3.87 seconds for the Eagle to hit the surface of the moon were it to free fall. The kinetic energy the Eagle would possess were it to fall from 40 feet is given by the simple equation mass x gravity x height and in this case that would be 15,200kg X 9.81 m/s per s/6 x 12.192 meters = 302,996 joules.

Lots of numbers, yet you left out the important one. What energy was absorbed/dissipated by the four struts on the landing pads?

Why did you not include that?

So if we take Armstrong at his word, a 40 foot Eagle fall under lunar conditions would mean the 15,200 kg bird would strike the lunar surface after a 3.87 second fall at a velocity of 6.33 meters per second or equivalently 20.6 feet per second or equivalently 13.6 miles and hour. The Eagle's kinetic energy on impact would be 302,996 joules.

False. You have not accounted for the struts absorbtion/dissipation.

I'll translate that into more familiar terms. The average compact car weighs roughly 3000-4,500 lbs. I'll use the halfway point, 3,750 lbs(1704 kg) The Eagle weighs 33,440 lbs without "gas" so that's 8.92 times as much as the compact car. I'll show an equivalent earth based situation using the compact car as the falling object, equivalent in that at the time the compact hits the earth it will be carrying 302,996 joules of energy.

False. You still have not accounted for the absorbtion/dissipation of the struts.

Here on earth

They were not on earth.

the kinetic energy of a falling object will of course also be given by mass x gravity x height. One wants here the height of a compact car fall that will give a solution providing 302,996 joules of energy. 302,996 divided by 1704 divided by 9.8 = 18.14 meters or 59 feet. So if a compact car were to be dropped from 59 feet, it would take 3.48 seconds to hit the earth with the same kinetic energy as Neil Armstrong's Eagle falling from 40 feet on the moon. My compact car would be moving at 34 meters per second at the time of impact. That translates to 112 feet per second or 76 miles an hour.

Still missing the absorbtion/dispertion of the struts.

So let's think about Armstrong's claim. He says that if he were to be able to control the Eagle's speed and attitude so that it simply dropped straight down, the thing could land on its legs and everything would be hunky dory.

It was designed to do exactly that, drop unpowered to the surface.

Doesn't make any sense at all does it now? Put those lunar module legs on a compact car. Secure them any way you like. Run the car into a wall at 76 miles per hour. And, given Armstrong's scenario, what's the likelihood under those circumstances that the thing will hit with one and not 4 legs. A 3750 lb car moving at 76 miles an hour running an Apollo lunar module leg into a solid piece of rock.

Now you are claiming Armstrong was not competent. Evidence please.

So Armstrong is way wrong here, either that or the dude is stranded on the moon with Aldrin.

Irrelevancy noted.

What's going on? Well it is just like when Collins says at the Apollo 11 post flight press conference that he couldn't recall seeing any stars when they traveled to the moon as the moon was eclipsing the sun. Same thing here. It is a ridiculous statement by Armstrong. He rarely messed up, but this here indeed was one of those rare screw ups. This of course is why Armstrong did not give interviews as a rule, did not write books, did not answer questions. The more of that stuff you do, the more this kind of stuff happens.

40 foot drop of a lunar module could easily break the leg on a lunar module and strand you with Aldrin.

Please provide references as to the strength of the LM struts on which you based your work.

 

[Jack by the hedge]

Wow. That wasn't worth the wait. Patrick's come back to the thread to demonstrate his incomprehesion of mechanics too.

A lunar lander hitting the moon at 13.6 mph may very well have the same KE as a car hitting a wall at 76 mph, but that does not mean that the latter event is in any way a useful model of the former.

 

[Patrick1000]

It should be obvious that it is not serious.....

Wow. That wasn't worth the wait. Patrick's come back to the thread to demonstrate his incomprehesion of mechanics too.

A lunar lander hitting the moon at 13.6 mph may very well have the same KE as a car hitting a wall at 76 mph, but that does not mean that the latter event is in any way a useful model of the former.

Scientific satire, a Karelesque reference to Jay's comment above.......

Need I spell everything out for you Jack by the hedge?

 

[JayUtah]

I have brought this subject up more than once before Jay, about Armstrong avoiding interviews and so forth to not get caught making such a gaffe as that referenced above.

Whitehouse's "biography" is not authorized. He didn't interview Armstrong.

Hansen's biography, on the other hand, is authorized. He interviewed Armstrong extensively. Eric Jones also interviewed Armstrong extensively. Further, Armstrong wasn't the only man on the spaceship. Aldrin does many, many publicity events. People obsess over Armstrong. That's how we know they're not historians.

Of course I am a doctor.

Why should anyone believe anything you say when you flip-flop so effortlessly among several contradictory claims? You have a reputation for saying anything that fits your claim at the moment. As soon as someone traced your photo to a real identity, you scrambled and said, "No, that's not me." Real professionals don't play games like this. Either you're a doctor and are willing to put your actual professional reputation behind your medical claims, or you are not a doctor and your opinion on Borman carries no weight.

Choose.

We should get back to debating the FACTS of Apollo Jay.

We can do that as soon as you stop posting something more than supposition and ignorant assumption and trying to pretend that it's fact.

And i am entitled to introduce any subject material relevant that I choose.

True, but you are not entitled to undeserved credibility after ignoring the responses to subject matter that you previously posted. Changing the subject in the middle of some line of questioning is evasion. You can behave as you wish, but you don't get to dictate what people think and say about the way you behave.

My point about the Armstrong gaffe above is a most excellent one, as are all of my other points as well.

No, it's a quote of dubious authenticity followed by a completely inept attempt to do physics. After such a convincing demonstration of your ineptitude, are you still trying to tell everyone you're a skilled engineer?

 

[Patrick1000]

Come on fess, thinbk a little less concretely here.....

Did they run out of fuel? Did they fall forty feet? If not, the rest of your post is meaningless.

Scientific satire, Karel, a double entendres without a risque interpretation....

If you do take it seriously, well then by all means , have at it, would be curious to see your defense of Armstrong.......

 

[Patrick1000]

Admittedly I did nbot double check, I used mgh for potential energy....

Go back and get a refund for your physics and maths degrees Pat - they taught you the wrong formula for deriving kinetic energy.

Kinetic energy is derived from the formula e=1/2m*v² where m is the mass in kg and v is the velocity in metres per second.

Using your velocity for the drop of the lunar lander, we get

e = 7600* 40.0689 = 304,523.64 joules of energy

Now your compact car, the speed of impact is a lot different from what you've derived:

In this case we're solving for v, so your formula is

v² = e/(1/2 m)

or

v = square root of the energy divided by one half the mass

using your mass for the car, I derive a velocity of 13.366 m/s or 29.83178 miles per hour.

All this of course is an exercise in remembering my high school physics from 1986 and pointless unless I knew what is the level of energy that the legs of the lander were designed to withstand?

Admittedly I did nbot double check, I used mgh for potential energy...and then solved for the little car's velocity using that. I'll check it again. I could easily have gotten it wrong. I typically don't use a calculator, not that that should matter necessarily one way or the other.

 

[matt.tansy]

Here on earth the kinetic energy of a falling object will of course also be given by mass x gravity x height. One wants here the height of a compact car fall that will give a solution providing 302,996 joules of energy. 302,996 divided by 1704 divided by 9.8 = 18.14 meters or 59 feet. So if a compact car were to be dropped from 59 feet, it would take 3.48 seconds to hit the earth with the same kinetic energy as Neil Armstrong's Eagle falling from 40 feet on the moon. My compact car would be moving at 34 meters per second at the time of impact. That translates to 112 feet per second or 76 miles an hour.

I love it when CTs try math. It always guarantees a laugh.

You screwed it up, son.

Here's a hint:

kinetic energy = mass ·velocity² · .5

Here's another hint:

one of your equations involves square roots that you failed to apply.

Try again!

After you learn sixth grade math and correct your errors then calculate how much deceleration the LM would have experienced if it dropped unpowered from 40 feet assuming the legs absorbed the impact by compressing 1 foot.

And, given Armstrong's scenario, what's the likelihood under those circumstances that the thing will hit with one and not 4 legs.

The likelihood is pretty darn good since he till had plenty of reaction control fuel remaining. Reaction control. Remember that? Reaction control? It's the thing you didn't know the LM had.

 

[Patrick1000]

I'll get right on it abadon, have the Thomas Kelly book here with me now along with .

Lots of numbers, yet you left out the important one. What energy was absorbed/dissipated by the four struts on the landing pads?

Why did you not include that?


False. You have not accounted for the struts absorbtion/dissipation.


False. You still have not accounted for the absorbtion/dissipation of the struts.


They were not on earth.


Still missing the absorbtion/dispertion of the struts.



It was designed to do exactly that, drop unpowered to the surface.



Now you are claiming Armstrong was not competent. Evidence please.


Irrelevancy noted.




Please provide references as to the strength of the LM struts on which you based your work.

I'll get right on it abadon, have the Thomas Kelly book here with me now along with more stats on the lander than you could shake a fraudulent moon walking stick at.

 

[matt.tansy]

Go back and get a refund for your physics and maths degrees Pat - they taught you the wrong formula for deriving kinetic energy.

Kinetic energy is derived from the formula e=1/2m*v² where m is the mass in kg and v is the velocity in metres per second.

Using your velocity for the drop of the lunar lander, we get

e = 7600* 40.0689 = 304,523.64 joules of energy

Now your compact car, the speed of impact is a lot different from what you've derived:

In this case we're solving for v, so your formula is

v² = e/(1/2 m)

or

v = square root of the energy divided by one half the mass

Gahhhh! Don't give him the answer! I love to watch CTs struggle with math!

 

[matt.tansy]

Why a compact car and not a boeing 747? What happens if we use a 747 with a maximum takeoff weight of 333,390 Kg?

Here on earth the kinetic energy of a falling object will of course also be given by mass x gravity x height. One wants here the height of a Boeing 747 fall that will give a solution providing 302,996 joules of energy. 302,996 divided by 333,390 divided by 9.8 = 0.09 meters or 3.5 inches.

So you can't land a 747 from 3 1/2 inches.

I was going to use an 80,000 ton aircraft carrier for my example?

 

[Patrick1000]

I see where I made my mistake, see if you can find yours.....

Go back and get a refund for your physics and maths degrees Pat - they taught you the wrong formula for deriving kinetic energy.

Kinetic energy is derived from the formula e=1/2m*v² where m is the mass in kg and v is the velocity in metres per second.

Using your velocity for the drop of the lunar lander, we get

e = 7600* 40.0689 = 304,523.64 joules of energy

Now your compact car, the speed of impact is a lot different from what you've derived:

In this case we're solving for v, so your formula is

v² = e/(1/2 m)

or

v = square root of the energy divided by one half the mass

using your mass for the car, I derive a velocity of 13.366 m/s or 29.83178 miles per hour.

All this of course is an exercise in remembering my high school physics from 1986 and pointless unless I knew what is the level of energy that the legs of the lander were designed to withstand?

I solved the problem by solving for the time of the fall. I came up with a fall of roughly 59, 60 feet. Then when I went to solve for the time of that fall, I used 9.8 meters per sec per sec for acceleration and not 32 feet per second squared so my time came out roughly twice what it should have bee. The correct time of fall for the compact is roughly 1.9 seconds and the velocity of course is given by the square root of (2gh), or equivalently as I solved it acceleration X time or 32 x 1.9 gives 60.8 feet per second or 41.5 miles per hour give or take.

Obviously just solving for the square root of 2gh is the easiest most straight forward way, but I was looking to provide all of the details of the event, time of fall and so forth.

Thanks for pointing that out. Your first answer is also incorrect as I am sure you have by now realized.

 

[nomuse]

Bzzzt! Wrong. Thanks for playing.

Why is he multiplying acceleration by distance as if it were time?

 

[Jack by the hedge]

...The lunar lander's mass was 15,200 kg without the propellant...

Not according to Wikipedia, it wasn't. That's approximately the mass they give for the lander with the descent propulsion system's fuel.

At 8,200kg, that fuel accounts for more than half the lander's entire mass. So if the lander ran out of fuel on descent, it would then be reduced to 6,834kg plus two astronauts. Not fifteen tons. Barely seven.

This is one small step into Patrick's calculation, one giant error already.