A question of credibility

[unknown]

I have seen again carefully the image of the first foot-print on the moon.
I'm really right: light is on the right, in fact all the ground around projects shadows on the left.
THE FOOT-PRINT IS IN ALTO-RILIEVO.
What a poor figure for Nasa. ;D ;D ;D

[JayUtah]

Mathematics is a rigorous discipline...

Yes. That's why I used it to support my argument.

When you apply mathematics to physics you can't consider all the various angles (because you don't know which they are)...

Hogwash. Physics is exactly the mathematical quanitification of the natural world. You can't do physics without math. Of course I know what all "the angles" are; it's my job.

...and you simplify the big complexity of problems...

No. We realize that the problem of flight dynamics is really the same simple planar problem repeated for each of three dimensions.

Be quiet, little man, you know almost nothing about forces that reign in the universe.

Refute my argument mathematically. Then you can call me names.

If you don't understand this simple thing, you are a big idiot...

Make up your mind. I'm either a "little man" or a "big idiot". I can't be both.

Here are three problems for you to solve to prove to us you really know about flight dynamics. If you decline to solve them, this conversation (and likely your sojourn here) is over.

1. Consider a spheroid spacecraft with a diameter of 3 meters hovering 100 meters above the lunar surface. The spacecraft has a gross mass of 1,000 kg. Assume for this problem that it has the mass properties of a solid sphere of the given mass. At the bottom of this spacecraft is an engine with a thrust of 100,000 N. The throat of the rocket is on the perimeter of the spacecraft sphere. The thrust -- for whatever reason -- goes 2.5 degrees off-axis. What is the rotation rate of the spacecraft in the off-axis plane after 1 second?

2. Solve the problem above with a thrust of 10,000 N.

3. Solve the problem above with a thrust of 10,000 N, but this time an altitude of only 50 meters.

These are extremely elementary problems in flight dynamics. If you can't solve these, then there is no reason to believe you have any basis from which to criticize anyone else. My staff engineers can solve these problems in about 5 minutes. As a control, I gave this problem to a 16-year-old high school physics student, who solved it in about 10 minutes. (He had to find some formulas.)

You have 24 hours. Put up or shut up.

[martin]

jayutah said:

These are extremely elementary problems in flight dynamics. If you can't solve these, then there is no reason to believe you have any basis from which to criticize anyone else. My staff engineers can solve these problems in about 5 minutes. As a control, I gave this problem to a 16-year-old high school physics student, who solved it in about 10 minutes. (He had to find some formulas.)

You have 24 hours. Put up or shut up.

I sent to you my answers by private message. But I think it does not matter, probably unknown is leaving us soon...

Martin

[JayUtah]

Martin got it right.

[unknown]

I wrote:
"When you apply mathematics to physics you can't consider all the various angles (because you don't know which they are)..."

JayUtah wrote:
"Of course I know what all "the angles" are; it's my job".

YOU ARE ONLY REALLY ARROGANT: also doctors do their job, but they have not yet found how to treat and to cure many kinds of cancer.

[Obviousman]

unknown said:

I wrote:
"When you apply mathematics to physics you can't consider all the various angles (because you don't know which they are)..."

JayUtah wrote:
"Of course I know what all "the angles" are; it's my job".

YOU ARE ONLY REALLY ARROGANT: also doctors do their job, but they have not yet found how to treat and to cure many kinds of cancer.

This guy is such a doofus.

Maybe we could keep him around as the court jester?

I think his name needs changing; instead of "unknown" it would be more like "without a clue".

[martin]

jayutah said:

Martin got it right.

And I needed 35 minutes, so I am 3.5 times slower than high school physics student But I have some excuses, it is many years since I have studied these things, I have derived the formula without a book, and people become slower when they are older...but intelligent person like unknown can solve this problem very quickly, I am sure.

datacable said:

"Earth diameter is 7,900 miles, and Moon diameter is 2,160 miles. It takes on average 90 minutes to complete one Earth orbit, so one Moon orbit should take roughly 25 minutes." - NasaScam

So by NasaScam it is possible to orbit the moon with the time I need to solve the problem of JayUtah, but I think this is not correct. I am calculating that time of orbit near surface of body is a function only of density of that object, and moon is less dense than earth, so low moon orbit takes longer than low earth orbit. So I think it is not possible to orbit the moon faster than I solve this problem. Am I correct?

Martin

[martin]

unknown said:

YOU ARE ONLY REALLY ARROGANT: also doctors do their job, but they have not yet found how to treat and to cure many kinds of cancer.

Can they cure you?

Martin

[JayUtah]

YOU ARE ONLY REALLY ARROGANT

Nope. It is proper for experts to speak from a position of authority. You are the layman trying to speak from a position of authority, but clearly faking it. That's what is arrogant. Arrogance is unwarranted confidence.

Now that you've posted in this thread, I know you've seen my challenge. The clock is ticking. Answer my questions correctly within 24 hours or leave. Time for you to put up or shut up.

[RAF]

jayutah said:

You have 24 hours. Put up or shut up.

Somehow, I picture Jay standing in the middle of main street of an old western town. Six shooters at the ready...and wearing the white cowboy hat, of course.

[martin]

raf said:

jayutah said:

You have 24 hours. Put up or shut up.

Somehow, I picture Jay standing in the middle of main street of an old western town. Six shooters at the ready...and wearing the white cowboy hat, of course.

So is unknown il brutto or il cattivo? ;D ;D ;D

Martin

[sts60]

Yes, doctors can't cure cancer all the time. So? Structural engineers can't build a bridge across the Atlantic Ocean. Do you deny the existence of bridges?

Interplanetary spaceflight (and lunar spaceflight by inclusion) is a well- bounded problem. "All" the angles are known for such missions. It's not like trying to fly to the far side of our galaxy. Your comparison is invalid.

[Bob B.]

martin said:

datacable said:

"Earth diameter is 7,900 miles, and Moon diameter is 2,160 miles. It takes on average 90 minutes to complete one Earth orbit, so one Moon orbit should take roughly 25 minutes." - NasaScam

So by NasaScam it is possible to orbit the moon with the time I need to solve the problem of JayUtah, but I think this is not correct. I am calculating that time of orbit near surface of body is a function only of density of that object, and moon is less dense than earth, so low moon orbit takes longer than low earth orbit. So I think it is not possible to orbit the moon faster than I solve this problem. Am I correct?

It a function of mass and distance. Here's the formula:

P^2 = 4*pi^2*r^3/GM

where,
p= period
r = radius of orbit
G = constant of gravitation
M = mass

It is usually easier the express GM as a constant, which for the Moon equals 4.902794E+12 m^3/s^2.

[martin]

bob said:

martin said:

So by NasaScam it is possible to orbit the moon with the time I need to solve the problem of JayUtah, but I think this is not correct. I am calculating that time of orbit near surface of body is a function only of density of that object, and moon is less dense than earth, so low moon orbit takes longer than low earth orbit. So I think it is not possible to orbit the moon faster than I solve this problem. Am I correct?

It a function of mass and distance. Here's the formula:

P^2 = 4*pi^2*r^3/GM

where,
p= period
r = radius of orbit
G = constant of gravitation
M = mass

It is usually easier the express GM as a constant, which for the Moon equals 4.902794E+12 m^3/s^2.

Yes, but mass is 4/3*pi*(r^3)*d where d is density where r is radius of this object. And if the orbit is close to surface, then this r is very close to r from above. So then period is:

P^2 = 3*pi*/(Gd)

This is my meaning.

Martin

[unknown]

JeyUtah in Reply #112 wrote:
"No, unknown, it is highly rude to make others try to guess at what you're thinking. If you have a question or a conclusion -- state it!"

Martin in Reply #110 wrote:
"No, I did not play your game. You were humourous for a short time, but this is ended".

I conduct the game, not you. ;D ;D ;D

I repeat:
I have seen again carefully the image of the first foot-print on the moon.
I'm really right: light is on the right, in fact all the ground around projects shadows on the left.
THE FOOT-PRINT IS IN ALTO-RILIEVO.
What a poor figure for Nasa. ;D ;D ;D