A question of credibility

[LunarOrbit]

Man, I wish I understood that stuff... I'm pretty good with basic math but algebra etc. makes my brain hurt.

[Bob B.]

martin said:

bob said:

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

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

Martin

Of course I can't prove I had my answer before the deadline. You'll just have to take my word for it.

[JayUtah]

I'll take something that's late and correct rather than on-time crap. The point of the exercise was to show that it's within the grasp of mere mortals to compute the elements of flight dynamics, although not necessarily within the grasp of blustery conspiracy theorists. It's important to note that those with expertise and experience (or at least willingness!) can solve the problem using less time than it took Unknown to complain about having to solve it. This resistance at all costs is characteristic of the troll, or the knowing charlatan.

[JayUtah]

Man, I wish I understood that stuff... I'm pretty good with basic math but algebra etc. makes my brain hurt.

Mathematical notation is indispensible for expressing proper relationships. But it's often detrimental for learning mathematics or the concepts behind it. I fear a lot of people are intimidated unnecessarily by strange symbols and seemingly arbitrary rules.

Algebra, of course, is what makes mathematics useful because it lets us solve problems symbolically, thereby letting us solve classes of problems instead of individual cases. Calculus and other "functional algebra" techniques let us find shortcuts to heinous problems, again in symbolic form.

But in this mad rush for uniformity and generality, the connection to the real world is often lost. When I was in college I took linear algebra from the mathematics department. Linear algebra is the study of vectors and associated concepts -- very useful for engineers. Unfortunately mathematicians seem excited by how abstract they can make something, not how helpful. I spent an entire semester dealing with vectors, and the professor once never drew an arrow on the board with a little number next to it: the classical representation of a "vector" in engineering. Engineers view vectors most often as directed quantities that relate to physical behavior.

Yes, of course you need to know the abstract theory. But it's rarely useful to introduce a concept as abstract theory. If you can express a concept correctly and clearly with just a few graphical strokes of the chalk, better than line after line of contrived gobbledy-gook, then you will have created understanding. This is why I usually explain the same concept two or three times, using different forms of communication.

Mathematicians may shudder at having a vector represented as an arrow, but it's often easier to make the leap from an intuitive approximation to raw theory than to try to get there by crunching the theory itself.

We're all familiar with Newton's

F = ma

which is, perhaps, better understood as

a = F/m

Basically, the heavier something is, the more you have to shove it in order to get it moving. That's all it's trying to express.

The leap of understanding is the realization that there are rotational equivalents to these concepts. Getting something turning is a matter of applying a turning force and being met with a degree of resistance that is particular to the object -- some notion of force and some notion of resistance.

The notion of force comes from the intuition of the wrench. If you want to open a fire hydrant, you attach the wrench and pull. If it doesn't budge, you either pull harder or get a longer wrench. The turning force depends on the linear force (how hard you pull) and the distance at which the force acts (the wrench length). That relationship suggests multiplication.

Unknown is right about one thing -- there are an infinite number of planes in which a rotational force can act. And so we have a convention that takes care of that. The "moment" or rotation force is a vector. The magnitude of the vector is the magnitude of the rotational force -- the length times the linear force. And the direction of the vector is the arrow pointing along the axis of the rotation. In the case of a fire hydrant that opens by turning anticlockwise, the direction vector points up. We use the "right-hand rule" to remember this. Hold your right hand so that your fingers curl in the direction of the rotation and your thumb points in the conventional direction of the vector. This clever convention lets us use vector arithmetic in order to sort out the various actual rotations that may be occurring in the object. There could be one, two, or eight million such vectors, and we could still compute the resulting "cardinal" rotations.

So much for the notion of force.

The notion of resistance is also pretty intuitive. Carry a ladder or a long piece of lumber and you know that getting it turning is hard. And more important, getting it to stop turning is hard too. A three-meter piece of steel plumbing pipe is hard to turn, harder than a three-meter piece of polyvinyl chloride pipe. Clearly heavier objects have more resistance to turning than lighter ones. Pure Newton. But 5 kilograms of PVC pipe is harder to turn than 5 kilograms of sugar in a bag. Shape also has something to do with it.

A light should be going off. Rotational force has a force component and a geometrical component. Rotational resistance has a mass component and a geometrical component. F = ma strikes again, only with geometry added.

Intuitively we can guess that for some constituent "particle" in an object, the heavier the particle, the more resistance it offers. And the farther that particle is away from the axis of rotation, the more resistance it offers. So the overall resistance of any body is the sum of the effects of all those particles. You could approximate the resistance by actually doing that addition using very small chunks of the object. But finding the cumulative effect of small parts of a problem -- actually showing what happens when those individual particles become infinitesimally small -- is the job of calculus. And so for bodies that have regular shapes with mathematically "pure" properties, we can devise formulas that describe their resistance symbolically, where you plug in only the raw mass and some information about the geometry (the length and diameter of a cylinder, the radius of a sphere, the dimensions of a slab, etc.). When you have an irregular object like a custom bearing, you have to fall back to the iterative method, subdividing the object into smaller and smaller "particles" and measuring their distance away from the center of rotation.

Trigonometry is one of the consummately useful branches of mathematics. And in our problem it gives us the "decomposition" of the off-axis thrust into the parts that help us determine rotation. It can be done purely in vector arithmetic too.

[martin]

jayutah said:

Unfortunately mathematicians seem excited by how abstract they can make something, not how helpful. I spent an entire semester dealing with vectors, and the professor once never drew an arrow on the board with a little number next to it: the classical representation of a "vector" in engineering. Engineers view vectors most often as directed quantities that relate to physical behavior.

Let us say only, different people undersand things in different ways ;D ;D ;D I am often solving problems in spaces of infinite dimensions; I do not know if the representation of vectors from physics or engineering as arrows is so useful in this case And many ideas of intuition are false in infinite dimensional case...

Martin

[Data Cable]

unknown said:

Helicopters don't fly because they create a low and a high pressure. They fly screwing in the air.

No, that's the Mile High Club... though I don't know if membership has ever been earned via helicopter.

[sts60]

*snork*
I'm sure. There are some pretty roomy helicopters out there...


The math comments remind me of Dave Barry's experience in calculus class (I may not have it exactly right):
"The professor would stand in front of the class and say, 'Consider the problem of a helix uncoiling in n dimensions...' He never said why it was a problem, or why we should worry about it if it was."

[bush]

JayUtah wrote:
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.

Mathematics has no importance if you don't know mechanics of the forces. In fact you are only big idiots and you never went to the moon and to Mars with your useless theories.

[papageno]

bush said:

JayUtah wrote
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.

Mathematics has no importance if you don't know mechanics of the forces. In fact you are only big idiots and you never went to the moon and to Mars with your useless theories.

Just fixed the quote tags.

[JayUtah]

Mathematics has no importance if you don't know mechanics of the forces.

Hi, "Unknown". You're about to be banned again, and probably reported to your ISP.

As it has been painfully made obvious, I do know the "mechanics of forces". And you were given ample opportunity to prove that you did too, but you found it more fun to hurl insults, such as:

In fact you are only big idiots...

You can't solve mechanics problems that even children can do correctly. You not only know less about the "mechanics of forces" than me or martin or Bob B., you know less about it than average students. Instead of insight and understanding, you offer only repetition and pathetic insults. You're substandard even for a layman.

[Bob B.]

bush said:

Mathematics has no importance if you don't know mechanics of the forces. In fact you are only big idiots and you never went to the moon and to Mars with your useless theories.

The above quote sounds too much like unknown to be a coincidence. I feel another ban coming on.

[sts60]

OK, "bush". You're calling people "idiots" implies you know better.

In that case, provide the solution to the problem. Show us where we're wrong. If you can't, then you clearly don't know better.

.
.
.

What's the matter? Cat got your tongue?

[martin]

bush said:

Mathematics has no importance if you don't know mechanics of the forces.

This is true! You can demonstrate to us superior knowledge of mechanics of forces by helicopter experiment I suggest to you. When you have anything which is important to do in your life, do it before though, because you will not have a chance later.

bush said:

In fact you are only big idiots and you never went to the moon and to Mars with your useless theories.

Mr. Vittorio Ferrari,

Why does a person who is 37 years old have no thing better to do than behave like a buffoon on the internet? Maybe you can go to a tavern in Italy, and talk to people in the way you talk to us. They will teach you proper manners very quickly.

Martin

[JayUtah]

When I lived in Italy I was driving with some Italian friends near the port of Catania. Ahead of us were some American sailors in a car, driving fast and swerving. At one point they threw a beer bottle out of the window, which smashed against the sidewalk near some pedestrians. Thankfully my friends knew that not all Americans behaved that way, but I was ashamed nonetheless.

Unknown, you disgrace yourself and your countrymen by behaving as you do. How are my colleagues here to discern your behavior from that of the average Italian, who is thoughtful, kind, and hospitable? You're doing the intellectual equivalent of throwing beer bottles at pedestrians. Che vergogna.

[papageno]

jayutah said:

Unknown, you disgrace yourself and your countrymen by behaving as you do. How are my colleagues here to discern your behavior from that of the average Italian, who is thoughtful, kind, and hospitable? You're doing the intellectual equivalent of throwing beer bottles at pedestrians. Che vergogna.

I do feel ashamed of this guy.

[rant]
Unfortunately I am not very surprised.
"Intellectuals"* have a traditional contempt for the hard sciences: they can proudly brag that they do not understand anything about mathematics, and ridicule scientists.
That is one of the reasons that there is no real political drive to support scientific research in Italy (there is a real brain drain).
[/rant]



* Not that I consider bush/unknown an intellectual.