A question of credibility

[martin]

jayutah said:

Tick tock, tick tock. Solve the simple problem and prove to us how superior you are.

He will not solve it. He will only say a lot of craps.

Martin

[martin]

unknown said:

My challenge is this:
Tell Nasa to build this old crock

Maybe you are too busy saying a lot of craps, so you forget my challenge. You will show to us stability of a helicopter without ability to change the angle of each rotor as it spins. Remember you can win a special award from this:

www.darwinawards.com

Martin

[JayUtah]

Since anyone can claim to be anything, or to be able to do anything, mere claims do not interest me. Many claims to expertise, such as Unknown's, can be demonstrated. And without the demonstration, the claim stands empty. He has said on numerous occasions that I don't know the behavior of natural forces. That implies he does. If those two propositions were true, he could describe how and why my explanations fail. That is, instead of just saying "Jay is wrong," he could say, "Jay is wrong because..." But aside from his repeated incorrect reference to balancing a bottle, he can't do it.

Unknown must supply some demostration of his claims to superior understanding. Otherwise it's just all talk.

[martin]

jayutah said:

Many claims to expertise, such as Unknown's, can be demonstrated.

I think unknown has demonstrated to every one already, what is his level of expertise ;D ;D ;D

But, if my guess is right, in some hours, he is out of this forum, and we do not listen to him say a lot of craps any more.

Martin

[unknown]

martin wrote:
"Maybe you are too busy saying a lot of craps, so you forget my challenge. You will show to us stability of a helicopter without ability to change the angle of each rotor as it spins. Remember you can win a special award from this:

www.darwinawards"

You talked about low pressure and high pressure of the rotor.
You talk about low pressure and high pressure also when you speak about wings of planes.

You have not understood anything about helicopters and planes.
How can a plane fly with a symmetrical shape of wings?
Because low pressure and high pressure of wings have no importance and it flys in spite of your wrong theories.

Helicopters don't fly because they create a low and a high pressure.
They fly screwing in the air. To be able to change the angle of their rotor make them able to fly slowly or fastly using better the power of the engine. ;D ;D ;D

[unknown]

JayUtah wrote:
"... He (unknown) has said on numerous occasions that I don't know the behavior of natural forces. That implies he does. If those two propositions were true, he could describe how and why my explanations fail. That is, instead of just saying "Jay is wrong," he could say, "Jay is wrong because..."

Because, if you were less arrogant, would know you make a big simplification to solve the problems of balance in the space: there are infinite moments of the forces in the space and you, instead, reduce them on a plane. So lunar modul doesn't work as you think. ;D ;D ;D

[Bob B.]

jayutah said:

Many claims to expertise, such as Unknown's, can be demonstrated. And without the demonstration, the claim stands empty.

It is also my experience that people who really can do this stuff are always quick and willing to demonstrate it. Remember Interdimensional Warrior at GLP? Whenever we challenged his knowledge on a subject it was always something like, "I don't need to prove myself to you", or "I'll provide the calculations at a time of my choosing". These responses are completely out of character for the engineering type personality.

[JayUtah]

Because, if you were less arrogant...

To employ one's legitimately obtained expertise is not arrogance. Arrogance is your attempts to pass off layman's knowledge as expert.

...would know you make a big simplification to solve the problems of balance in the space

But you can't even solve the very simplest problems. So you have no business trying to guess at the complexity of the practical solutions. You can't even discuss the situation in which only one moment acts.

...there are infinite moments of the forces in the space

And Chapter One of any dynamics textbook is how to reckon them in the three axes!

Moments by definition are planar and cannot be otherwise. Yes, there may be several arbitrary planes involved, but the first step is to be able to solve the singular planar problem. Then we can move on to discuss how those moments are decomposed in the orthogonal vector basis.

So lunar modul doesn't work as you think.

It works exactly as I think. And you're out of time.

Sorry, you've failed the test. You've proved you're a troll. Goodbye.

[JayUtah]

How can a plane fly with a symmetrical shape of wings?

Most airfoils do not have symmetrical cross sections.

But to answer your question, a symmetrical cross section can be set an at angle of attack that creates a downdraft at the rear of the wing. This produces not only direct lift, but it sets up a counter-rotation of air around the wing -- air "wants" to rotate such that it passes forward along the bottom and rearward along the top. This slows the air passing under the wing and accelerates the air passing above it, generating the classic zones of pressure difference.

Because low pressure and high pressure of wings have no importance and it flys in spite of your wrong theories.

False. Although the common flat-bottomed, round-topped airfoil model in elementary textbooks is impractical, the principle of airfoil flight still relies on generating higher pressure below the wing than above it.

Helicopters don't fly because they create a low and a high pressure.

Then where does the dust come from when they're hovering?

[martin]

jayutah said:

Sorry, you've failed the test. You've proved you're a troll. Goodbye.

I think every one knows this already for a long time...

Martin

[JayUtah]

And now the answer.

Martin got the right answers, but he didn't show his work.

First the mass properties of our spacecraft. Of course in actual practice, real spacecraft mass properties are detailed. Nice synthetic shapes like spheres and cylinders have mass properties we can compute analytically. Arbitrary shapes are computed iteratively. Thankfully there are some theorems that let us build up mass properties for entire vehicles by considering the properties of their constituent parts and knowing where they sit in the combined coordinate space of the vehicle.

So I threw him a bone by allowing him to use a classic synthetic shape (a solid sphere) as the basis for understanding the mass properties of the spacecraft.

We are concerned with the spacecraft's center of mass and its moment of inertia around that center of mass. In a solid sphere the center of mass is located at the geometrical center of the sphere. The moment of inertia for a solid sphere is given by the formula

I = 2/5 M R^2

where M is the scalar mass in kilograms and R is the radius of the sphere in meters. Here,

I = (2/5)(1,000 kg)(1.5 m)^2
I = 900 kg m^2

That's the "resistance" of a sphere of that mass and geometry to rotational forces (moments).

Now compute the moment. This requires us to compute the length of the moment arm and the portion of the off-center thrust that is perpendicular to that moment arm.

The throat of the rocket is on the spacecraft perimeter, which is just another way of saying that the moment arm length for the main engine here is equivalent to the radius of the spacecraft: 1.5 meters.

By directing the engine off-axis by 2.5 degrees, we define a right triangle of characteristic angle 2.5 degrees and a hypotenuse equal to the magnitude of the thrust. The right-angle edges of the triangle define how much of the thrust is along the axis and how much of it is perpendicular to the axis. The axial thrust will determine linear propulsion. The transverse thrust defines the rotational moment.

The transverse component Fx is therefore given by

Fx = sin(Err) * Ft

Where Err is the angle of engine alignment "error" and Ft is total thrust.

Fx = sin(2.5) * 100,000 N
Fx = 4,362 N

This thrust magnitude, multiplied by the length of the moment arm, gives the rotational moment.

mx = Fx * L
mx = (4,362 N)(1.5 m)
mx = 6,543 N m

In order to arrive at a rotational acceleration, this moment must be divided by the moment of inertia, according to the law of conservation of rotational momentum. This is equivalent to dividing force by mass in the linear case in order to discover linear acceleration.

d(w)/d(t) = mx / I

where d(w) denotes the change in angular velocity (i.e., d[theta]/d[t]) and d(t) denotes the change in time, and mx and I are as above.

Since we normally compute these for each cardinal plane, and since there is a shorthand notation for "rate of change with respect to time," these components can sometimes be thought of as "x-dot" "y-dot" and "z-dot" in three dimensions. Since we're not specifically in any of these planes, d(w)/d(t) suffices.

d(w)/d(t) = (6,543 N m) / (900 kg m^2)
d(w)/d(t) = 7.27 radians / s

In practical terms that is a very significant rotation, suggesting that 100,000 N is far too powerful a motor for this spacecraft.

Recomputing for 10,000 N is simple. Since all our computations that depend on thrust have been linear, we can simply adjust the answer by the ratio of prior thrust to new thrust. That is, 10,000 N is one-tenth the thrust of 100,000 N and so the rotation rate is one-tenth: 0.727 radians / s.

Note that we have completely disregarded the information given for altitude above the lunar surface. That is because such altitude, and whatever gravity might exist either at a fixed distance above the lunar surface or in open space, is utterly irrelevant. Any forces of gravity that apply to our spacecraft will apply at its center of mass, and thus will generate no rotational moments to muddy up these computations. The moment arm in all cases is zero.

Thus the answer to the third part is the same as the second part because it is, in every respect, an identical computation.

Having obtained the rotational acceleration, normally we integrate over time in order to find a rotation rate. The time interval was given here as one second, so the integration is trivial.

[martin]

jayutah said:

And now the answer.

Martin got the right answers, but he didn't show his work.

I can fax to you, but it is long ;D ;D ;D I could not remember what is the energy of a spinning sphere, so I have to calculate a triple integral. Then I determine how much force generates sideways motion and how much spinning motion by decomposition of forces and arguments of symmetry. Then I calculate rotation rate by conservation of energy argument. I think your method is easier...

Martin

[JayUtah]

Yes, my method is much easier. But your having computed it via a different method and having arrived at the same answer is an important check. For real engineering we often do this, but we always show work so that it can be reviewed later if needed.

[Bob B.]

Here's my calculations. Took me 5 minutes at most.

[martin]

bob said:

Here's my calculations. Took me 5 minutes at most.

OK, you can win first prize. I will win second only

Martin