a thought
- Thread starter ruthless
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ruthless
Rift Surfer
from the article: "The simplest operation that can be performed on a vector is to multiply it by a scalar. This scalar multiplication alters the magnitude of the vector. In other word, it makes the vector longer or shorter. "
ok, a few questions. is this simply a visual form of the math behind it? is this simple addition and subtraction? a+b=c, but what excactly is c? and how do i calculate it?
sorry in advance for any dumb questions. /ttiforum/images/graemlins/smile.gif
ok, a few questions. is this simply a visual form of the math behind it? is this simple addition and subtraction? a+b=c, but what excactly is c? and how do i calculate it?
sorry in advance for any dumb questions. /ttiforum/images/graemlins/smile.gif
ruthless
Rift Surfer
another quote from the article: "Don't be intimidated by vectors. When you're first introduced to them, it can seem like they're overwhelming, but some effort and attention to detail will result in quickly mastering the concepts involved."
thats an understatement! i find it very intimidating, and foreign. i do find it similar to the xy coordinates i was taught in the sixth grade though.
i am very distracted at the moment, im trying to do quite a few things at once. when it settles down tonight, i will go back and reread everything in peace. i think i kind of understand, but missing some things. i'll get into it a little more when i get my quiet time tonight. /ttiforum/images/graemlins/smile.gif
thats an understatement! i find it very intimidating, and foreign. i do find it similar to the xy coordinates i was taught in the sixth grade though.
i am very distracted at the moment, im trying to do quite a few things at once. when it settles down tonight, i will go back and reread everything in peace. i think i kind of understand, but missing some things. i'll get into it a little more when i get my quiet time tonight. /ttiforum/images/graemlins/smile.gif
RainmanTime
Super Moderator
ruthless,
You are asking questions and that is good!! Here is something else that is good:
That is a big-time connection and we should run with it. In fact, the concept of directionality (x, y, and also z!) is half of what vectors are all about. You will always hear people say that vectors have magnitude and direction. Force is a good vector quantity to think about because we are all familiar with the magnitude of a force...for example our weight. I am about 175 pounds. That is the magnitude of my force that my body exerts on the earth. But what is its direction? Well, thanks to what we know about gravity it is straight down towards the center of the earth. But a vector, such as a force, can point in ANY direction. Think of an aircraft engine that has thrust vectoring technology. For a conventional engine the thrust always points straight out the tailpipe. By adding variable geometry (motors controlling the shape of the nozzle) we can cause the thrust vector to point in a bunch of different directions that may NOT be straight out the tailpipe. We could point it up or down or to the left or right of straight out the tail pipe.
So since space is 3-dimensional we need to be able to express the direction of a vector in all 3 spatial dimensions. This is where coordinate systems come into play...We will typially set up an X-Y-Z coordinate system and then break down any vector into its orthogonal components within these 3 dimensions. This is also where trigonometry comes into play, and it is one fundamental form of mathematics that vector math is built upon. We can deal with that as we go. Now on to other comments/questions:
No. A and B are vectors, just as they are pictured there. The best visualization for a vector is a line with an arrow at the end. The length of the line represents the magnitude of the vector (how big it is). The direction the arrow is pointing represents...well, its direction. The main thing we must realize is a simple geometric reality. ANY two vectors (lines), no matter how they are oriented, ALWAYS define a 2-D plane. Or stated alternately, I can always draw a plane that will contain any two vectors.
C is the vector that results when we mutiply those two vectors (using the vector cross product). We denote this in an equation as:
C = A x B
Note we use "x" and here it means "vector cross product". When we use "*" it represents the scalar (dot) product. One principle of the vector cross product that we can prove with geometry and trigonometry is that the result of crossing any two vectors (which form a plane) gives a vector (in this case C) that is NORMAL TO (perpendicular to) the plane that contains the two vectors being crossed. As I tried to explain to Einstein there is a very easy way to visualize this with your right hand. Hold your right hand out so your fingers extend flat and your thumb is pointing up. In this orientation your fingers pointing out straight represent the direction of vector A. As you now curl your fingers inward to form a fist you are mimicing the operation of "crossing A into B". (Imagine the vector B as pointing outward from your palm). Now, the direction that your thumb is pointing represents the direction of the C vector that results from crossing A into [/b]B[/b]
This is where trigonometry comes into play. Have you ever taken a class in trigonometry, ruthless? The first line is merely showing us how we determine the angle between a vector F and the X and Y axes of the coordinate system. Using the theorm of Pythagorous we can determine that the trigonometric cosine of the angle between the vector F and the X axis is calculated by dividing the magnitude of the X component of the F force by the total magnitude of the F. Same thing for the sin of the angle, except that it uses the Y component of the F force. The second line has simply used algebra to solve the first two equations for the Fx and Fy components.
No. This applet is only to help you visualize how any two vectors, when multiplied with the vector cross priduct, result in a vector that is perpendicular to the plane that contains the first two vectors. But this concept (the vector cross product) applies to a great many things in physics! It is a key to knowledge!
There are no dumb questions. The answer to the first question is "yes", kind of. If you mean the applet, the answer is "yes, definitely". The applet allows you to physically see two vectors, and the plane that is defined by them...and allows you to play around with the lengths and orientations of the A and B vectors such that you can see what happens to the C vectpr as a result of varying conditions for A and B. The answer to your second question is "no". We are not adding vectors, we are multiplying them.
The quote you repeated above was talking about the scalar (dot) product. This is the more common form of multiplication you are used to. All it does is to either lengthen or shorten the magnitude of a vector...but it does NOT change the vector's pointing direction. Now, where the vector cross product is different is that it actually multiplies the magnitudes of the two vectors and it also combines the two directions of the two source vectors to resolve the final direction of the final vector. This is what makes the cross product so unique and important! It addresses the fact that vectors pointing in space combine in spatial manners....in addition to simply growing or shrinking (the vector dot product).
The simplest vector cross product to understand is the torque.
T = r x F
r is the vector that points from the central point of rotation to the point where the force F is applied.
F is the force itself that is applied about the central point to create the torque T
Again you can use your right hand to visualize this. Hold it out flat again with your thumb pointing up. The distance from your thumb to the end of your fingers represents the vector r. If you not visualize a force F pushing on the ends of your fingers to force them to curl up, you are simulating the vector cross product of r x F. The result of this force applied at this distance from the central rotating point (your thumb) results in a TORQUE which is the rotational equivalent of a force. That torque is represented by your thumb. And by the "right hand rule" convention we say this is a positive torque. If we were to perform the opposite cross product multiplication of F x r we would get the same magnitude of the torque, but it would be pointing in the OPPOSITE direction.
I hope some of this is helping you understand.
RMT
You are asking questions and that is good!! Here is something else that is good:
That is a big-time connection and we should run with it. In fact, the concept of directionality (x, y, and also z!) is half of what vectors are all about. You will always hear people say that vectors have magnitude and direction. Force is a good vector quantity to think about because we are all familiar with the magnitude of a force...for example our weight. I am about 175 pounds. That is the magnitude of my force that my body exerts on the earth. But what is its direction? Well, thanks to what we know about gravity it is straight down towards the center of the earth. But a vector, such as a force, can point in ANY direction. Think of an aircraft engine that has thrust vectoring technology. For a conventional engine the thrust always points straight out the tailpipe. By adding variable geometry (motors controlling the shape of the nozzle) we can cause the thrust vector to point in a bunch of different directions that may NOT be straight out the tailpipe. We could point it up or down or to the left or right of straight out the tail pipe.
So since space is 3-dimensional we need to be able to express the direction of a vector in all 3 spatial dimensions. This is where coordinate systems come into play...We will typially set up an X-Y-Z coordinate system and then break down any vector into its orthogonal components within these 3 dimensions. This is also where trigonometry comes into play, and it is one fundamental form of mathematics that vector math is built upon. We can deal with that as we go. Now on to other comments/questions:
No. A and B are vectors, just as they are pictured there. The best visualization for a vector is a line with an arrow at the end. The length of the line represents the magnitude of the vector (how big it is). The direction the arrow is pointing represents...well, its direction. The main thing we must realize is a simple geometric reality. ANY two vectors (lines), no matter how they are oriented, ALWAYS define a 2-D plane. Or stated alternately, I can always draw a plane that will contain any two vectors.
C is the vector that results when we mutiply those two vectors (using the vector cross product). We denote this in an equation as:
C = A x B
Note we use "x" and here it means "vector cross product". When we use "*" it represents the scalar (dot) product. One principle of the vector cross product that we can prove with geometry and trigonometry is that the result of crossing any two vectors (which form a plane) gives a vector (in this case C) that is NORMAL TO (perpendicular to) the plane that contains the two vectors being crossed. As I tried to explain to Einstein there is a very easy way to visualize this with your right hand. Hold your right hand out so your fingers extend flat and your thumb is pointing up. In this orientation your fingers pointing out straight represent the direction of vector A. As you now curl your fingers inward to form a fist you are mimicing the operation of "crossing A into B". (Imagine the vector B as pointing outward from your palm). Now, the direction that your thumb is pointing represents the direction of the C vector that results from crossing A into [/b]B[/b]
This is where trigonometry comes into play. Have you ever taken a class in trigonometry, ruthless? The first line is merely showing us how we determine the angle between a vector F and the X and Y axes of the coordinate system. Using the theorm of Pythagorous we can determine that the trigonometric cosine of the angle between the vector F and the X axis is calculated by dividing the magnitude of the X component of the F force by the total magnitude of the F. Same thing for the sin of the angle, except that it uses the Y component of the F force. The second line has simply used algebra to solve the first two equations for the Fx and Fy components.
No. This applet is only to help you visualize how any two vectors, when multiplied with the vector cross priduct, result in a vector that is perpendicular to the plane that contains the first two vectors. But this concept (the vector cross product) applies to a great many things in physics! It is a key to knowledge!
There are no dumb questions. The answer to the first question is "yes", kind of. If you mean the applet, the answer is "yes, definitely". The applet allows you to physically see two vectors, and the plane that is defined by them...and allows you to play around with the lengths and orientations of the A and B vectors such that you can see what happens to the C vectpr as a result of varying conditions for A and B. The answer to your second question is "no". We are not adding vectors, we are multiplying them.
The quote you repeated above was talking about the scalar (dot) product. This is the more common form of multiplication you are used to. All it does is to either lengthen or shorten the magnitude of a vector...but it does NOT change the vector's pointing direction. Now, where the vector cross product is different is that it actually multiplies the magnitudes of the two vectors and it also combines the two directions of the two source vectors to resolve the final direction of the final vector. This is what makes the cross product so unique and important! It addresses the fact that vectors pointing in space combine in spatial manners....in addition to simply growing or shrinking (the vector dot product).
The simplest vector cross product to understand is the torque.
T = r x F
r is the vector that points from the central point of rotation to the point where the force F is applied.
F is the force itself that is applied about the central point to create the torque T
Again you can use your right hand to visualize this. Hold it out flat again with your thumb pointing up. The distance from your thumb to the end of your fingers represents the vector r. If you not visualize a force F pushing on the ends of your fingers to force them to curl up, you are simulating the vector cross product of r x F. The result of this force applied at this distance from the central rotating point (your thumb) results in a TORQUE which is the rotational equivalent of a force. That torque is represented by your thumb. And by the "right hand rule" convention we say this is a positive torque. If we were to perform the opposite cross product multiplication of F x r we would get the same magnitude of the torque, but it would be pointing in the OPPOSITE direction.
I hope some of this is helping you understand.
RMT
ruthless
Rift Surfer
interesting indeed. i have a small understanding of it, but am missing alot. no, ive never taken trigonometry or algebra. my experience in math is limited to addition, subtraction, multiplication, and division unfortunately.
it is my opinion that this is what i need to do: go back to school and get a better education. i think that before i even attempt to try to do these advanced things, i need to build a good foundation in mathematics. i think it makes things alot harder for me if im just going to try to learn on the fly. yes, im very intimidated by this stuff at this time, funny as that is. im going to attempt to get a good understanding of all things in it at the same time, but, its going to be hard. i didnt understand half of the things on that site.
"No. A and B are vectors, just as they are pictured there. The best visualization for a vector is a line with an arrow at the end. The length of the line represents the magnitude of the vector (how big it is). The direction the arrow is pointing represents...well, its direction. The main thing we must realize is a simple geometric reality. ANY two vectors (lines), no matter how they are oriented, ALWAYS define a 2-D plane. Or stated alternately, I can always draw a plane that will contain any two vectors.
C is the vector that results when we mutiply those two vectors (using the vector cross product). We denote this in an equation as:
C = A x B"
ok, heres my problem. im not understanding this because, well, im just not getting it. so lets see if i can ask a few questions and figure this one out.
"Force is a good vector quantity to think about because we are all familiar with the magnitude of a force...for example our weight. I am about 175 pounds. That is the magnitude of my force that my body exerts on the earth. But what is its direction? Well, thanks to what we know about gravity it is straight down towards the center of the earth. But a vector, such as a force, can point in ANY direction. Think of an aircraft engine that has thrust vectoring technology. For a conventional engine the thrust always points straight out the tailpipe. By adding variable geometry (motors controlling the shape of the nozzle) we can cause the thrust vector to point in a bunch of different directions that may NOT be straight out the tailpipe. We could point it up or down or to the left or right of straight out the tail pipe."
ok, could we set up an example. lets use the aircraft engine for example. the force it applies is the vector (a)? and (b) is gravity or drag maybe? and what exactly does (c) calculate?
sorry ray, im having a tough time with this one.
it is my opinion that this is what i need to do: go back to school and get a better education. i think that before i even attempt to try to do these advanced things, i need to build a good foundation in mathematics. i think it makes things alot harder for me if im just going to try to learn on the fly. yes, im very intimidated by this stuff at this time, funny as that is. im going to attempt to get a good understanding of all things in it at the same time, but, its going to be hard. i didnt understand half of the things on that site.
"No. A and B are vectors, just as they are pictured there. The best visualization for a vector is a line with an arrow at the end. The length of the line represents the magnitude of the vector (how big it is). The direction the arrow is pointing represents...well, its direction. The main thing we must realize is a simple geometric reality. ANY two vectors (lines), no matter how they are oriented, ALWAYS define a 2-D plane. Or stated alternately, I can always draw a plane that will contain any two vectors.
C is the vector that results when we mutiply those two vectors (using the vector cross product). We denote this in an equation as:
C = A x B"
ok, heres my problem. im not understanding this because, well, im just not getting it. so lets see if i can ask a few questions and figure this one out.
"Force is a good vector quantity to think about because we are all familiar with the magnitude of a force...for example our weight. I am about 175 pounds. That is the magnitude of my force that my body exerts on the earth. But what is its direction? Well, thanks to what we know about gravity it is straight down towards the center of the earth. But a vector, such as a force, can point in ANY direction. Think of an aircraft engine that has thrust vectoring technology. For a conventional engine the thrust always points straight out the tailpipe. By adding variable geometry (motors controlling the shape of the nozzle) we can cause the thrust vector to point in a bunch of different directions that may NOT be straight out the tailpipe. We could point it up or down or to the left or right of straight out the tail pipe."
ok, could we set up an example. lets use the aircraft engine for example. the force it applies is the vector (a)? and (b) is gravity or drag maybe? and what exactly does (c) calculate?
sorry ray, im having a tough time with this one.
ruthless
Rift Surfer
i just went back and looked at the java applet again. if a is thrust and b is gravity, is c the net thrust? i think im incorrect, because when i replace b gravity with drag, it does not seem to work. if i put b at a 180 degree angle from a, c becomes nothing, no matter the magnitude of a or b.
ive gotta figure this out or i swear, my brain will explode... /ttiforum/images/graemlins/confused.gif
ive gotta figure this out or i swear, my brain will explode... /ttiforum/images/graemlins/confused.gif
Einstein
Dimensional Traveler
ruthless
I can sympathize with you. I started to read through the link that Ray gave me for the gyroscope. Every fourth word I had to look up. Only problem is the word was described with words I was unfamiliar with. So this went on for about an hour. At which time I noticed I was just finishing the first paragraph. So I decided to take a break. I haven't gotten back to it yet. So it's highly unlikely I will completely comprehend it that way. However I do comprehend Ray's links on vectors and the vector cross product. The way I comprehend it is the vector cross product is a mathematical construct that extends the product of two vectors, which exist in two dimensional space, into the third dimension. So it is an artificially created 3-D vector. I suspect it is being used to describe phenomena that apppears to be 3-D based. I'm still studying the concept. I don't know if it's legal to do it that way. But as far as I know the legislators haven't passed any laws restricting the path mathematicians decide to take. I'm sure that will change once they get wind of it.
I can sympathize with you. I started to read through the link that Ray gave me for the gyroscope. Every fourth word I had to look up. Only problem is the word was described with words I was unfamiliar with. So this went on for about an hour. At which time I noticed I was just finishing the first paragraph. So I decided to take a break. I haven't gotten back to it yet. So it's highly unlikely I will completely comprehend it that way. However I do comprehend Ray's links on vectors and the vector cross product. The way I comprehend it is the vector cross product is a mathematical construct that extends the product of two vectors, which exist in two dimensional space, into the third dimension. So it is an artificially created 3-D vector. I suspect it is being used to describe phenomena that apppears to be 3-D based. I'm still studying the concept. I don't know if it's legal to do it that way. But as far as I know the legislators haven't passed any laws restricting the path mathematicians decide to take. I'm sure that will change once they get wind of it.
RainmanTime
Super Moderator
ruthless,
Whatever you do, keep doing what you are doing and do NOT fall down the hole that Einstein has chosen!!
You ALREADY ARE figuring it out. Little by little. And it is NOT an easy concept that you are tackling, especially having not taken any trigonometry courses. That makes it especially difficult, but not impossible.
First I want you to note one thing: You are stretching for physical interpretations (that is the way of the engineer!) which is good. But note also that you are seeking to understand what amounts to rotational dynamics with linear force concepts only (thrust, drag, and weight due to gravity are all linear force concepts). The world of rotational kinematics (which is a term that groups statics and dynamics together) is wonderful, POWERFUL, and a bit daunting when you first decide to study it. But trust me, the payback is immense. Do not give up stalking, oh warrior! :D
Second, I want you to congratulate yourself on a discovery you have made that will increase your understanding of rotational kinematics. Namely, You have found something interesting about when vectors A and B are at 180 degrees from each other... and a similar thing occurs when they are coincident with each other (0 degree angle between them). The resultant vector C becomes ZERO!
Now let me frame this in a physical situation so you can see what this means and why it is an important realization: Let's say I have a completely solid bowling ball (no holes drilled in it yet). We are going to talk torques here. The r vector for this torque will be from the geometric center of the bowling ball (which is also its center of mass) out to the edge of the ball. Now let's say that I apply a force to try and lift the ball and the direction of that force is either 0 degrees with respect to the r vector, or 180 degrees with it. Essentially what this means is that I am lifting the ball (applying a force) directly through its center of mass. Imagine me pulling vertically upward on a string through the center of the ball Will the ball REVOLVE (i.e. torque) at all? NO! Hence, the torque applied to the ball by applying the force for lifting at that angle (0 degrees angle between r and F) EQUALS ZERO.
But now let us apply the force F such that there is a 90 degree angle between this force and the vector r. This would amount to the ball sitting on the ground and the vector r being oriented horizontally (parallel with the ground). We now tie a string to the side of the ball as it sits on the ground (to the end of the r vector) and we exert an upward force F to pull on the ball. Draw this picture for yourself and convince yourself that the direction you are first applying the force F is 90 degrees (perpendicular) to the orientation of the vector r. And now what happens to the ball when you apply that force and lift the ball off the ground??? Specifically, what sort of MOTION does the ball exhibit when you do this???
YES...IT TORQUES!! The ball rotates. By applying the force at some angle OTHER than 0 or 180 degrees, you have been able to induce rotational motion. Hence, the resultant vector (the torque) is NOT ZERO!!
Again, congratulate yourself ruthless, because in your actions to understand, you have hit upon a fact that you did not understand (the 180 degrees thing) and I have now explained why this makes PHYSICAL SENSE. This is the best way to learn engineering (study, then ask questions, and get answers!)
You are making progress!
RMT
Whatever you do, keep doing what you are doing and do NOT fall down the hole that Einstein has chosen!!
You ALREADY ARE figuring it out. Little by little. And it is NOT an easy concept that you are tackling, especially having not taken any trigonometry courses. That makes it especially difficult, but not impossible.
First I want you to note one thing: You are stretching for physical interpretations (that is the way of the engineer!) which is good. But note also that you are seeking to understand what amounts to rotational dynamics with linear force concepts only (thrust, drag, and weight due to gravity are all linear force concepts). The world of rotational kinematics (which is a term that groups statics and dynamics together) is wonderful, POWERFUL, and a bit daunting when you first decide to study it. But trust me, the payback is immense. Do not give up stalking, oh warrior! :D
Second, I want you to congratulate yourself on a discovery you have made that will increase your understanding of rotational kinematics. Namely, You have found something interesting about when vectors A and B are at 180 degrees from each other... and a similar thing occurs when they are coincident with each other (0 degree angle between them). The resultant vector C becomes ZERO!
Now let me frame this in a physical situation so you can see what this means and why it is an important realization: Let's say I have a completely solid bowling ball (no holes drilled in it yet). We are going to talk torques here. The r vector for this torque will be from the geometric center of the bowling ball (which is also its center of mass) out to the edge of the ball. Now let's say that I apply a force to try and lift the ball and the direction of that force is either 0 degrees with respect to the r vector, or 180 degrees with it. Essentially what this means is that I am lifting the ball (applying a force) directly through its center of mass. Imagine me pulling vertically upward on a string through the center of the ball Will the ball REVOLVE (i.e. torque) at all? NO! Hence, the torque applied to the ball by applying the force for lifting at that angle (0 degrees angle between r and F) EQUALS ZERO.
But now let us apply the force F such that there is a 90 degree angle between this force and the vector r. This would amount to the ball sitting on the ground and the vector r being oriented horizontally (parallel with the ground). We now tie a string to the side of the ball as it sits on the ground (to the end of the r vector) and we exert an upward force F to pull on the ball. Draw this picture for yourself and convince yourself that the direction you are first applying the force F is 90 degrees (perpendicular) to the orientation of the vector r. And now what happens to the ball when you apply that force and lift the ball off the ground??? Specifically, what sort of MOTION does the ball exhibit when you do this???
YES...IT TORQUES!! The ball rotates. By applying the force at some angle OTHER than 0 or 180 degrees, you have been able to induce rotational motion. Hence, the resultant vector (the torque) is NOT ZERO!!
Again, congratulate yourself ruthless, because in your actions to understand, you have hit upon a fact that you did not understand (the 180 degrees thing) and I have now explained why this makes PHYSICAL SENSE. This is the best way to learn engineering (study, then ask questions, and get answers!)
You are making progress!
RMT
RainmanTime
Super Moderator
It is a difficult concept. But the more you study it, and definitely the more you apply it, the clearer it will become. I guarantee you that someday you will look back on this moment and think "how could I have NOT gotten it? It is so simple!" /ttiforum/images/graemlins/smile.gif This happens all the time with my students.
Here is another good link to study up on vectors:
http://www.euclideanspace.com/maths/algebra/vectors/index.htm
In this link you will begin to see their relationship to matrices and matrix mathematics. There are a lot of other good links to follow there to gain more information.
RMT
RainmanTime
Super Moderator
Einstein,
You are now coming to an understanding why getting a degree in science (and moreso in engineering) is difficult, and why people who get them are prized individuals (i.e. people pay them good money to come work for them). You are getting a peak into a complex world, but rest assured (and I know you have your doubts) it is a highly productive world, where we regularly use math that you think is useless to accurately describe reality.
I am now going to ask you to apply the "visualization" process that you are always ranting and raving to me about. The cross product is not an artificially created 3-D vector. It has a basis in physical reality. The visualization I want you to perform is the one I just wrote up for ruthless. The bowling ball and applying a force to the external surface of the ball with two different orientations. If the cross product (in this case, torque) of the r and the F vector was "artificial" then it would not create the reality of the physical motion of the ball torquing (rotating) when the force F is applied at a right angle to the r vector. So finally we see an application of reality through visualization where you cannot squirm away from me. Here we have a physical indication that the vector cross product (in this case the torque) is real. The orientation of the cross product vector is telling us what axis the body will rotate around due to the applied torque. The length of the cross product vector tells us the magnitude of that overall torque that is rotating the ball!
Once we agree and understand this topic, then we can move on to discussing Angular Momentum (also a vector which results from a cross product), and the conservation thereof.
RMT
You are now coming to an understanding why getting a degree in science (and moreso in engineering) is difficult, and why people who get them are prized individuals (i.e. people pay them good money to come work for them). You are getting a peak into a complex world, but rest assured (and I know you have your doubts) it is a highly productive world, where we regularly use math that you think is useless to accurately describe reality.
I am now going to ask you to apply the "visualization" process that you are always ranting and raving to me about. The cross product is not an artificially created 3-D vector. It has a basis in physical reality. The visualization I want you to perform is the one I just wrote up for ruthless. The bowling ball and applying a force to the external surface of the ball with two different orientations. If the cross product (in this case, torque) of the r and the F vector was "artificial" then it would not create the reality of the physical motion of the ball torquing (rotating) when the force F is applied at a right angle to the r vector. So finally we see an application of reality through visualization where you cannot squirm away from me. Here we have a physical indication that the vector cross product (in this case the torque) is real. The orientation of the cross product vector is telling us what axis the body will rotate around due to the applied torque. The length of the cross product vector tells us the magnitude of that overall torque that is rotating the ball!
Once we agree and understand this topic, then we can move on to discussing Angular Momentum (also a vector which results from a cross product), and the conservation thereof.
RMT
Einstein
Dimensional Traveler
RMT
During the course of my studying the cross product, I came across an anomaly that I can't explain. The word torque seems to have aquired a different meaning than from when I learned it. I learned that torque is a turning force. But all the definitions that I access now show it as a turning force times the radius. And the vector for this force is along the axis that the turning force acts about. How can a vector exist on the axis if a force is applied at the radius in the direction of rotation about the axis? I might be able to do this by combining two or more forces, but not just one. Could you try another visualization? And pay special attention to not using the word torque, as it seems to confuse me more and more.
During the course of my studying the cross product, I came across an anomaly that I can't explain. The word torque seems to have aquired a different meaning than from when I learned it. I learned that torque is a turning force. But all the definitions that I access now show it as a turning force times the radius. And the vector for this force is along the axis that the turning force acts about. How can a vector exist on the axis if a force is applied at the radius in the direction of rotation about the axis? I might be able to do this by combining two or more forces, but not just one. Could you try another visualization? And pay special attention to not using the word torque, as it seems to confuse me more and more.
RainmanTime
Super Moderator
Einstein,
I am afraid that all I can say is that whoever taught you about a torque initially did a poor job. An alternate word for torque is moment. However, it is quite clear this is the appropriate definition (force times a distance) when you reference automotive specs when they tell you to "torque the nut to xxx inch-pounds". Now, a torque can cause rotation or turning, but it can also cause torsion (essentially the rotational component of stress).
The vector exists on that axis because that is the axis about which the combined vectors (i.e. radius x force) acts about. You need to come to grips with the fact that a torque (moment) is a combined metric (distance times force). That combination transforms the force vector (oriented in one direction) and the radius vector (oriented in some other direction) into a 3rd (combined) vector that operates "about an axis mutually orthogonal to the two original vectors". You must admit that the vector being along that axis does, indeed, indicate the axis about which rotation (or torsion) will manifest. This is the "direction" aspect of the torque vector. The "magnitude" aspect is represented by the length of the torque vector.
I don't understand this comment. The new vector (torque) is describing something different than a force...because it is a force acting at a distance. If you were to include two forces, acting at opposite sides of the rotation point but causing rotation in the same direction, we call that a "couple". (Two moments acting to cause motion or stress in the same direction are defined as a couple).
I will try to get back to you with a different visualization. But as for the use of the word torque, I am using it correctly. I could use the word moment, but they mean exactly the same thing. What I am suggesting is that this confusion by you could be "the problem" behind why you have had issues with rotational kinematics for a long time. Correct your understanding of what a torque really is (as opposed to what someone allowed you to believe it was) and you will be one step closer to understanding the accuracy of the mathematics.
I think the next visualization (not this post...later) will probably focus on angular momentum, as it is also a vector that describes rotational kinematics, it is also the result of a cross product, and it will be the next concept we need to understand gyro kinematics.
RMT
I am afraid that all I can say is that whoever taught you about a torque initially did a poor job. An alternate word for torque is moment. However, it is quite clear this is the appropriate definition (force times a distance) when you reference automotive specs when they tell you to "torque the nut to xxx inch-pounds". Now, a torque can cause rotation or turning, but it can also cause torsion (essentially the rotational component of stress).
The vector exists on that axis because that is the axis about which the combined vectors (i.e. radius x force) acts about. You need to come to grips with the fact that a torque (moment) is a combined metric (distance times force). That combination transforms the force vector (oriented in one direction) and the radius vector (oriented in some other direction) into a 3rd (combined) vector that operates "about an axis mutually orthogonal to the two original vectors". You must admit that the vector being along that axis does, indeed, indicate the axis about which rotation (or torsion) will manifest. This is the "direction" aspect of the torque vector. The "magnitude" aspect is represented by the length of the torque vector.
I don't understand this comment. The new vector (torque) is describing something different than a force...because it is a force acting at a distance. If you were to include two forces, acting at opposite sides of the rotation point but causing rotation in the same direction, we call that a "couple". (Two moments acting to cause motion or stress in the same direction are defined as a couple).
I will try to get back to you with a different visualization. But as for the use of the word torque, I am using it correctly. I could use the word moment, but they mean exactly the same thing. What I am suggesting is that this confusion by you could be "the problem" behind why you have had issues with rotational kinematics for a long time. Correct your understanding of what a torque really is (as opposed to what someone allowed you to believe it was) and you will be one step closer to understanding the accuracy of the mathematics.
I think the next visualization (not this post...later) will probably focus on angular momentum, as it is also a vector that describes rotational kinematics, it is also the result of a cross product, and it will be the next concept we need to understand gyro kinematics.
RMT
Einstein
Dimensional Traveler
RMT
I think you're moving too fast here. So I think maybe we should share our knowledge about rotational kinematics. Somewhere between the two of us there will be more discrepencies that could clear up some of my confusion. You have to remember that I actually test out alot of my knowledge that I aquire. So somewhere along the way I may have discovered something that is in disagreement with the math. That could also be some of the basis for my confusion as well. The confusion could also be in some of the basic definitions. Remember I said it was like the math behind the gyroscope had been rewritten. When I first came across the math, it was all in the form of radians. Another thing I notice is that the torque which equals force times radius comes out dimensionally as energy. So here goes: An out of balance mass rotating about an axis will rotate about the axis at a CONSTANT SPEED as long as a continuous rotational force is applied. The center of mass is no longer on the axis of rotation. It is similar to the gyroscope with one exception. With the out of balance mass all rotational forces are in the same plane. With the gyro, the applied rotational force is orthogonal to the precessional force. But the gyro will precess at a CONSTANT SPEED with the constant application of a rotational force. So logically that does suggest that the precessional rotational force could be due to a center of mass no longer residing on the axis of rotation.
Now, do you disagree with anything so far? Also the use of the term vector? I was always under the impression that a vector was a linear force. Is there such a thing as a curvilinear force vector? The difference I see between the two is that a straight vector has its direction along one dimension of length. But a curved vector occupies two dimensions. Ah ha! That helps. Torque actually has the extra dimension of length thus making it two dimensional. But I still don't see any way there can be a vector along the axis of rotation. Maybe it would help if you distinguish between single dimensional force vector and two dimensional force vector. Another thing is that to me, an applied torque is in the direction of rotation. Is that correct?
I think you're moving too fast here. So I think maybe we should share our knowledge about rotational kinematics. Somewhere between the two of us there will be more discrepencies that could clear up some of my confusion. You have to remember that I actually test out alot of my knowledge that I aquire. So somewhere along the way I may have discovered something that is in disagreement with the math. That could also be some of the basis for my confusion as well. The confusion could also be in some of the basic definitions. Remember I said it was like the math behind the gyroscope had been rewritten. When I first came across the math, it was all in the form of radians. Another thing I notice is that the torque which equals force times radius comes out dimensionally as energy. So here goes: An out of balance mass rotating about an axis will rotate about the axis at a CONSTANT SPEED as long as a continuous rotational force is applied. The center of mass is no longer on the axis of rotation. It is similar to the gyroscope with one exception. With the out of balance mass all rotational forces are in the same plane. With the gyro, the applied rotational force is orthogonal to the precessional force. But the gyro will precess at a CONSTANT SPEED with the constant application of a rotational force. So logically that does suggest that the precessional rotational force could be due to a center of mass no longer residing on the axis of rotation.
Now, do you disagree with anything so far? Also the use of the term vector? I was always under the impression that a vector was a linear force. Is there such a thing as a curvilinear force vector? The difference I see between the two is that a straight vector has its direction along one dimension of length. But a curved vector occupies two dimensions. Ah ha! That helps. Torque actually has the extra dimension of length thus making it two dimensional. But I still don't see any way there can be a vector along the axis of rotation. Maybe it would help if you distinguish between single dimensional force vector and two dimensional force vector. Another thing is that to me, an applied torque is in the direction of rotation. Is that correct?
RainmanTime
Super Moderator
Einstein,
Often have I commended you for this (the testing), but equally often I have pointed out where you come to conclusions based on insufficient measurement data (i.e. a video is not precision measurements). And when you have used a precision device like your accelerometer, I pointed out that "measuring" an effect that occurs well faster than the accelerometer's bandwidth has inherent sensor problems where you cannot guarantee what the sensor is reporting is actually going on due to the effect you are measuring only. While your zeal to do real experiments is wonderful, your instrumentation is most often much too meager to even build a math model of what you think you are seeing, much less come to the kinds of conclusions or theories you have come to.
So above you have put forth your two theories about why a gyro doesn't behave like classical equations say it does. What is interesting is that you seem to put forward these theories as if they are the most probable. (either you found some new aspect of physics no one else has observed, or someone changed the equations on you). When in reality the highest probability theory is that you simply do not understand the classical equations, and are holding out hope that they do not accurately describe reality... but they do, and this hase been proven over and over again.
What does that mean? Surely it cannot ALL be in radians, for they are merely a measure of angular displacement. Since there is clearly energy involved, and we know (kinetic) energy is proportional to mass and velocity squared, then clearly the units of mass, space and time must figure into any explanation of the gyro. Certainly a gyro's spin rate (angular velocity) could be expressed in terms of radians per second. But radians in and of themself cannot fully explain the rotational kinematics of gyros. It seems you wish to hold out hope that the gyro equations I have pointed you to are somehow incorrect...when they are not.
Yes, indeed. And do you know how energy and torque differ, even though their units are the same? Energy is a scalar (it has magnitude but no direction) whereas torque is the result of a cross product and as such has both magnitude AND direction.
In what direction must that continuous force be applied? The answer to this question shows why we need vectors. The direction you apply the force matters, because if the force is applied in the wrong direction you statement above will not be true...and it can be measured as such. What you are scratching to try and describe is the angular version of Newton's first law of rectilinear motion. You know: "A body in motion will remain in motion at a constant speed unless acted upon by an unbalanced force." This generalized truth of linear motion is also true of angular motion. This truth is what gives way to Newton's Second Law of Motion, which is also true both rectilinearly and rotationally. From this truth come the equations:
Force = Mass*Acceleration (rectilinear motion) .....
F=m*a
and
Moment (Torque) = Moment of Inertia * Angular Acceleration (rotational motion).....
M=I*alpha
This is the basic truth you must begin with to analyze the gyro motion, Einstein. Just as the web page does that I pointed you to. Do you not believe these accurately describe non-relativistic motion of bodies???
This is certainly not always true, because an out of balance mass has non-principal moments of inertia. What this means is for an unbalanced mass the cross products of inertia in the inertia tensor are non-zero. Here is what the inertia tensor looks like symbolically:
[Ixx...Ixy...Ixz]
[Iyx...Iyy...Iyz]
[Izx...Izy...Izz]
In a balanced mass every member of this tensor except for Ixx, Iyy, and Izz is zero. These are called the moments of inertia. The off-axis terms (Ixy, Ixz, Iyz, Iyz, Izx, and Izy) are called the products of inertia. When these values are non-zero, the mass is unbalanced about one or more axes. Again Einstein, this is all elementary and has been proven.
This is fully explained by the classical equations when you next consider the change in angular momentum of the spinning gyro when results from the applied torque (rotational force). So this bit of logic you use is equivalent to saying that the Precessional torque (rotational force) is equal to the time rate of change of angular momentum. Again, it is Newton's Second Law stated in terms of momentum. Nothing new here.
Some minor points, as stated above. But the bigger point is that this is all described with existing math based on Newtonian dynamics. It works. There is nothing new you have discovered.
A vector does not always have to be a force. For example, angular momentum is a vector, and momentum is not a force.
That is exactly what a torque (moment) is, Einstein. Curvilinear requires a radius to a point which represents the center of the rotational part of the curvilinear motion. That is precisely why a torque (moment) requires a cross product between the radial vector and the applied force vector. But again...they are all vectors Einstein.
Yes. And this length dimension (the radius) is one vector... the applied force is another vector. The two vectors taken together do define a 2-D plane. That axis of the rotation that results from the torque is normal to the plane defined by the radisu and force vectors. This is what the vector cross product means:
M = r x F
This is what you are trying to describe, Einstein. It is one in the same with the classical math. And again you are (hopefully) now seeing why the cross product is required, and why it helps describe the situation of rotational kinematics.
The final torque vector that results from the cross product is along the axis of rotation because that is describing the rotational motion that results from the torque! It describes the DIRECTION of the rotational motion...so of course it should be along that axis...that is, in fact, the axis that the whole thing is rotating about. The length of that vector describes the magnitude of the "curvilinear force" (as you call it). The distance and applied force define a plane. The resulting rotation is about the axis normal to that plane.
An applied force CAUSES a rotation. The vector cross product describes the axis about which that rotation takes place. The right hand rule handles the "directionality" in that the resulting c vector points in one direction (normal to the plane) for clockwise rotation. But it will point in the opposite direction (but still normal to the plane) for counter-clockwise rotation.
I am going to start another post to try and hammer this point home, Einstein. It will explain how we use rectilinear forces and curvilinear torques to fully model the kinematics of an airplane. Hopefully it will finally convince you that the existing math can describe the translational AND ROTATIONAL kinematics of any body quite accurately.
RMT
Often have I commended you for this (the testing), but equally often I have pointed out where you come to conclusions based on insufficient measurement data (i.e. a video is not precision measurements). And when you have used a precision device like your accelerometer, I pointed out that "measuring" an effect that occurs well faster than the accelerometer's bandwidth has inherent sensor problems where you cannot guarantee what the sensor is reporting is actually going on due to the effect you are measuring only. While your zeal to do real experiments is wonderful, your instrumentation is most often much too meager to even build a math model of what you think you are seeing, much less come to the kinds of conclusions or theories you have come to.
So above you have put forth your two theories about why a gyro doesn't behave like classical equations say it does. What is interesting is that you seem to put forward these theories as if they are the most probable. (either you found some new aspect of physics no one else has observed, or someone changed the equations on you). When in reality the highest probability theory is that you simply do not understand the classical equations, and are holding out hope that they do not accurately describe reality... but they do, and this hase been proven over and over again.
What does that mean? Surely it cannot ALL be in radians, for they are merely a measure of angular displacement. Since there is clearly energy involved, and we know (kinetic) energy is proportional to mass and velocity squared, then clearly the units of mass, space and time must figure into any explanation of the gyro. Certainly a gyro's spin rate (angular velocity) could be expressed in terms of radians per second. But radians in and of themself cannot fully explain the rotational kinematics of gyros. It seems you wish to hold out hope that the gyro equations I have pointed you to are somehow incorrect...when they are not.
Yes, indeed. And do you know how energy and torque differ, even though their units are the same? Energy is a scalar (it has magnitude but no direction) whereas torque is the result of a cross product and as such has both magnitude AND direction.
In what direction must that continuous force be applied? The answer to this question shows why we need vectors. The direction you apply the force matters, because if the force is applied in the wrong direction you statement above will not be true...and it can be measured as such. What you are scratching to try and describe is the angular version of Newton's first law of rectilinear motion. You know: "A body in motion will remain in motion at a constant speed unless acted upon by an unbalanced force." This generalized truth of linear motion is also true of angular motion. This truth is what gives way to Newton's Second Law of Motion, which is also true both rectilinearly and rotationally. From this truth come the equations:
Force = Mass*Acceleration (rectilinear motion) .....
F=m*a
and
Moment (Torque) = Moment of Inertia * Angular Acceleration (rotational motion).....
M=I*alpha
This is the basic truth you must begin with to analyze the gyro motion, Einstein. Just as the web page does that I pointed you to. Do you not believe these accurately describe non-relativistic motion of bodies???
This is certainly not always true, because an out of balance mass has non-principal moments of inertia. What this means is for an unbalanced mass the cross products of inertia in the inertia tensor are non-zero. Here is what the inertia tensor looks like symbolically:
[Ixx...Ixy...Ixz]
[Iyx...Iyy...Iyz]
[Izx...Izy...Izz]
In a balanced mass every member of this tensor except for Ixx, Iyy, and Izz is zero. These are called the moments of inertia. The off-axis terms (Ixy, Ixz, Iyz, Iyz, Izx, and Izy) are called the products of inertia. When these values are non-zero, the mass is unbalanced about one or more axes. Again Einstein, this is all elementary and has been proven.
This is fully explained by the classical equations when you next consider the change in angular momentum of the spinning gyro when results from the applied torque (rotational force). So this bit of logic you use is equivalent to saying that the Precessional torque (rotational force) is equal to the time rate of change of angular momentum. Again, it is Newton's Second Law stated in terms of momentum. Nothing new here.
Some minor points, as stated above. But the bigger point is that this is all described with existing math based on Newtonian dynamics. It works. There is nothing new you have discovered.
A vector does not always have to be a force. For example, angular momentum is a vector, and momentum is not a force.
That is exactly what a torque (moment) is, Einstein. Curvilinear requires a radius to a point which represents the center of the rotational part of the curvilinear motion. That is precisely why a torque (moment) requires a cross product between the radial vector and the applied force vector. But again...they are all vectors Einstein.
Yes. And this length dimension (the radius) is one vector... the applied force is another vector. The two vectors taken together do define a 2-D plane. That axis of the rotation that results from the torque is normal to the plane defined by the radisu and force vectors. This is what the vector cross product means:
M = r x F
This is what you are trying to describe, Einstein. It is one in the same with the classical math. And again you are (hopefully) now seeing why the cross product is required, and why it helps describe the situation of rotational kinematics.
The final torque vector that results from the cross product is along the axis of rotation because that is describing the rotational motion that results from the torque! It describes the DIRECTION of the rotational motion...so of course it should be along that axis...that is, in fact, the axis that the whole thing is rotating about. The length of that vector describes the magnitude of the "curvilinear force" (as you call it). The distance and applied force define a plane. The resulting rotation is about the axis normal to that plane.
An applied force CAUSES a rotation. The vector cross product describes the axis about which that rotation takes place. The right hand rule handles the "directionality" in that the resulting c vector points in one direction (normal to the plane) for clockwise rotation. But it will point in the opposite direction (but still normal to the plane) for counter-clockwise rotation.
I am going to start another post to try and hammer this point home, Einstein. It will explain how we use rectilinear forces and curvilinear torques to fully model the kinematics of an airplane. Hopefully it will finally convince you that the existing math can describe the translational AND ROTATIONAL kinematics of any body quite accurately.
RMT
RainmanTime
Super Moderator
What We Mean By: SIX DEGREES OF FREEDOM
Einstein,
You may have heard (or maybe not) that aircraft simulations are often called "Six Degree Of Freedom" simulations. We in the biz shorten that to say it is a 6-DOF equation. Well, what exactly ARE those "six degrees" of freedom and how do they relate to classical kinematics equations that accurately describe motion?
Quite simply, 3 of the degrees of freedom come from the expression of Newton's Second Law with respect to rectilinear motion. Newton's Second Law is a vector equation, and can be written in two ways:
<font color="red">F = m*a = d(m*V)/dt[/COLOR]
So any vector equation can be decomposed into its related set of 3 scalar equivalent equations. In other words, we can write equations for the X, Y, and Z motions in 3 separate equations:
<font color="red">Fx = m*ax = m*(dVx/dt) ---- For X-axis motion
Fy = m*ay = m*(dVy/dt) ---- For Y-axis motion
Fz = m*az = m*(dVz/dt) ---- For Z-axis motion[/COLOR]
<font color="blue"> We call these the THREE RECTILINEAR DEGREES OF FREEDOM OF MOTION.[/COLOR]
And the other 3 degrees of freedom that make up the full set of 6 comes from a similar treatment of Newton's Second Law for rotational motion. We write Newton's Second Law in terms of rotational kinematics as follows:
<font color="red">M = I*alpha = d(I*Omega)/dt[/COLOR]
alpha is the angular acceleration of the body.
omega is the angular velocity of the body.
I = inertia tensor
Again, this is a VECTOR equation just like its force equivalent. But it describes the rotational kinematics of the body, whereas the force equation describes the rectilinear kinematics of the body. We break this vector equation down into scalar rotations about each of the 3 axes:
<font color="red">Mx = Ixx*alphax = d(Ixx*Omegax)/dt ---- For rotation about X-axis
My = Iyy*alphay = d(Iyy*Omegay)/dt ---- For rotation about Y-axis
Mz = Izz*alphaz = d(Izz*Omegaz)/dt ---- For rotation about Z-axis[/COLOR]
<font color="blue"> We call these the THREE ROTATIONAL DEGREES OF FREEDOM OF MOTION.[/COLOR]
When we put the 3 rectilinear DOFs together with the 3 rotational DOFs in a simulation, what we get is often called:
THE SIX CURVILINEAR DEGREES OF FREEDOM OF AIRCRAFT MOTION.
Curvilinear (6 DOF) = Rectilinear (3 DOF) + Rotational (3 DOF)
Make sense, Einstein? And all of this is also used to describe the kinematics of gyros. It all "works" very well, as is shown by every flying vehicle we build. There is nothing "unknown" here that you may have found. Instead, there is something that has been unknown to you because, perhaps, you did not understand kinematics completely. Perhaps these discussions have helped you come to understand some of these equations, and why they work so well.
Do you now understand how all of this comes from nothing more than Newton's Second Law, applied to 3-dimensional space using vectors?
RMT
Einstein,
You may have heard (or maybe not) that aircraft simulations are often called "Six Degree Of Freedom" simulations. We in the biz shorten that to say it is a 6-DOF equation. Well, what exactly ARE those "six degrees" of freedom and how do they relate to classical kinematics equations that accurately describe motion?
Quite simply, 3 of the degrees of freedom come from the expression of Newton's Second Law with respect to rectilinear motion. Newton's Second Law is a vector equation, and can be written in two ways:
<font color="red">F = m*a = d(m*V)/dt[/COLOR]
So any vector equation can be decomposed into its related set of 3 scalar equivalent equations. In other words, we can write equations for the X, Y, and Z motions in 3 separate equations:
<font color="red">Fx = m*ax = m*(dVx/dt) ---- For X-axis motion
Fy = m*ay = m*(dVy/dt) ---- For Y-axis motion
Fz = m*az = m*(dVz/dt) ---- For Z-axis motion[/COLOR]
<font color="blue"> We call these the THREE RECTILINEAR DEGREES OF FREEDOM OF MOTION.[/COLOR]
And the other 3 degrees of freedom that make up the full set of 6 comes from a similar treatment of Newton's Second Law for rotational motion. We write Newton's Second Law in terms of rotational kinematics as follows:
<font color="red">M = I*alpha = d(I*Omega)/dt[/COLOR]
alpha is the angular acceleration of the body.
omega is the angular velocity of the body.
I = inertia tensor
Again, this is a VECTOR equation just like its force equivalent. But it describes the rotational kinematics of the body, whereas the force equation describes the rectilinear kinematics of the body. We break this vector equation down into scalar rotations about each of the 3 axes:
<font color="red">Mx = Ixx*alphax = d(Ixx*Omegax)/dt ---- For rotation about X-axis
My = Iyy*alphay = d(Iyy*Omegay)/dt ---- For rotation about Y-axis
Mz = Izz*alphaz = d(Izz*Omegaz)/dt ---- For rotation about Z-axis[/COLOR]
<font color="blue"> We call these the THREE ROTATIONAL DEGREES OF FREEDOM OF MOTION.[/COLOR]
When we put the 3 rectilinear DOFs together with the 3 rotational DOFs in a simulation, what we get is often called:
THE SIX CURVILINEAR DEGREES OF FREEDOM OF AIRCRAFT MOTION.
Curvilinear (6 DOF) = Rectilinear (3 DOF) + Rotational (3 DOF)
Make sense, Einstein? And all of this is also used to describe the kinematics of gyros. It all "works" very well, as is shown by every flying vehicle we build. There is nothing "unknown" here that you may have found. Instead, there is something that has been unknown to you because, perhaps, you did not understand kinematics completely. Perhaps these discussions have helped you come to understand some of these equations, and why they work so well.
Do you now understand how all of this comes from nothing more than Newton's Second Law, applied to 3-dimensional space using vectors?
RMT
ruthless
Rift Surfer
Re: What We Mean By: SIX DEGREES OF FREEDOM
i was walking through the house yesterday, and i saw a book that caught my eye. ive got no idea whose it was, but im happy i found it. its called, "basic technical mathematics with calculus 6th edition metric version."
it says in the book that its supposed to cover 4 semesters. i have a question. would this book give me pretty much all the information i would need to have a good foundation to start learning aerospace engineering?
i was walking through the house yesterday, and i saw a book that caught my eye. ive got no idea whose it was, but im happy i found it. its called, "basic technical mathematics with calculus 6th edition metric version."
it says in the book that its supposed to cover 4 semesters. i have a question. would this book give me pretty much all the information i would need to have a good foundation to start learning aerospace engineering?
RainmanTime
Super Moderator
Re: What We Mean By: SIX DEGREES OF FREEDOM
What an excellent coincidence, eh ruthless? Seems like it came at just the right time, huh?
It won't give you all the information you will need, but it will be one hell of a start! /ttiforum/images/graemlins/yum.gif It will surely introduce you to trigonometry, analytic geometry, derivative calculus, integral calculus, infinite series and may even get your started with differential equations. These topics I have just rattled off here are the fundamental basics of engineering math. The stuff we use the most. I am really glad you found this book. I can point you to topics from this math when they come up in our discussions.
One thing about aerospace engineering is that we take more math courses than any of the other engineering disciplines. While most engineering students can finish their math in two years (freshman and sophomore), an aerospace engineer still goes on to take at least 2, usually 3 more courses in advanced math beyond sophomore years.
The subject of that advanced math is exactly what i have been discussing here with you and Einstein... It is the application of VECTOR mathematics to all the basics calculus and differential equations you learn in the early calculus courses.
Vector and tensor math *IS* the defacto language of physics. Once you learn it (and we are forever trying to master it) you have gained the biggest tool in an engineer's toolbox.
RMT
What an excellent coincidence, eh ruthless? Seems like it came at just the right time, huh?
It won't give you all the information you will need, but it will be one hell of a start! /ttiforum/images/graemlins/yum.gif It will surely introduce you to trigonometry, analytic geometry, derivative calculus, integral calculus, infinite series and may even get your started with differential equations. These topics I have just rattled off here are the fundamental basics of engineering math. The stuff we use the most. I am really glad you found this book. I can point you to topics from this math when they come up in our discussions.
One thing about aerospace engineering is that we take more math courses than any of the other engineering disciplines. While most engineering students can finish their math in two years (freshman and sophomore), an aerospace engineer still goes on to take at least 2, usually 3 more courses in advanced math beyond sophomore years.
The subject of that advanced math is exactly what i have been discussing here with you and Einstein... It is the application of VECTOR mathematics to all the basics calculus and differential equations you learn in the early calculus courses.
Vector and tensor math *IS* the defacto language of physics. Once you learn it (and we are forever trying to master it) you have gained the biggest tool in an engineer's toolbox.
RMT
Einstein
Dimensional Traveler
RMT
Thanks for clearing that up for me. The 3-D visualization in the link on cross product was very confusing to me.
Let's play a little catch up before I address some of your other interesting points. I got so interested in the actual science behind an accelerometer, that I was able to construct one myself. I discovered that an accelerometer uses silicon as a piezoelectric material to generate a voltage signal in the presence of an acceleration. But there is also another way that silicon material can generate a voltage signal. In the presence of an electric field. A moving electric field will produce the same effect as an acceleration. OK, so maybe all my results are actually from the electric fields generated from my test equipment. More experimentation was in order to find out. I took a one megahertz silicon RF crystal and connected it to a scope lead. I rapidly moved the crystal up and down and watched for any voltage on the scope. YES! A crude accelerometer, but it does work. Next I encased the cyrstal in a grounded metal can to act as a faraday cage to shield it from the effects of any electric fields. At around 250 kilohertz I actually get my test equipment to cause my homemade accelerometer to generate voltage. It isn't much, about 10 miliivolts. But it isn't electric waves that are causing it. It is a resonant frequency of the crystal I'm using. I reasoned that I could move forward by using a broader frequency spectrum of crystals to cover more frequencies. Or I could just use an off the shelf integrated circuit and shield it in a metal can. But I stopped there. I'm missing something. How do I turn those miniscule gravity waves into a propulsion device? I'm very sure I can amplify the effect. But it was time to stop and gather more clues elsewhere.
As you may be aware, I've decided to revisit the gyroscope in my experimentation. I firmly believe that all of the answers that physics seeks lies within the framework of the gyroscope. As you've already seen, I have an alternate way to comprehend the gyroscope. Now that you have helped me over that basic understanding of cross product I would like to move on. Slowly of course. You mentioned something of great interest in a previous post. I think you called it a "couple". But today you mentioned something that I have been thinking about for that last three weeks. It was Degrees Of Freedom.
Well of course I disagree. You might have noticed that I have alternate interpretations. They may lead to the same result in the end. But the path I took is different. I see different conclusions along the way. The degrees of freedom that you talk about look like directions in time to me. I believe that would be something new. A more basic understanding of time and how it works. Of course if you keep instructing me on how to actually apply the math to this. Mainstream science might take notice. You know the math is just a fringe benefit to me. I'll probably have a time machine up and running before I fully develop the math for it. I see six degrees of freedom for torque. Two directions in each orthogonal plane of 3-D space. So you might be either referring to a plane or a line as a degree of freedom if I try to interpret what you say from context. I merely chose a direction as being a direction in time. One of those six directions is responcible for the forward flow of time. With the gyroscope it is only possible to access four out of the six. And only three at a time. The open ended orthogonal torque would occur in orthogonal time. Any transfer of force in orthogonal time would appear to be instantaneous. And that is what my current experiment is all about. You might like it. I'm coupling two gyroscopes together in an attempt to create two spiral force directions that cause the device to move in a more stabilized pattern. The double spiral could be described as a double helix. If it works as planned, then I might try building a device that accesses another direction of time.
As for the math, you caught my interest when you mentioned a "couple". I think that might be a mathematical area that would cover some of my current experimentation. But there appears to be more than one way to couple torques together.
This last month I got a mini mill from Harbor Frieght. This thing cuts steel like butter. I just love to fabricate stuff.
Thanks for clearing that up for me. The 3-D visualization in the link on cross product was very confusing to me.
Let's play a little catch up before I address some of your other interesting points. I got so interested in the actual science behind an accelerometer, that I was able to construct one myself. I discovered that an accelerometer uses silicon as a piezoelectric material to generate a voltage signal in the presence of an acceleration. But there is also another way that silicon material can generate a voltage signal. In the presence of an electric field. A moving electric field will produce the same effect as an acceleration. OK, so maybe all my results are actually from the electric fields generated from my test equipment. More experimentation was in order to find out. I took a one megahertz silicon RF crystal and connected it to a scope lead. I rapidly moved the crystal up and down and watched for any voltage on the scope. YES! A crude accelerometer, but it does work. Next I encased the cyrstal in a grounded metal can to act as a faraday cage to shield it from the effects of any electric fields. At around 250 kilohertz I actually get my test equipment to cause my homemade accelerometer to generate voltage. It isn't much, about 10 miliivolts. But it isn't electric waves that are causing it. It is a resonant frequency of the crystal I'm using. I reasoned that I could move forward by using a broader frequency spectrum of crystals to cover more frequencies. Or I could just use an off the shelf integrated circuit and shield it in a metal can. But I stopped there. I'm missing something. How do I turn those miniscule gravity waves into a propulsion device? I'm very sure I can amplify the effect. But it was time to stop and gather more clues elsewhere.
As you may be aware, I've decided to revisit the gyroscope in my experimentation. I firmly believe that all of the answers that physics seeks lies within the framework of the gyroscope. As you've already seen, I have an alternate way to comprehend the gyroscope. Now that you have helped me over that basic understanding of cross product I would like to move on. Slowly of course. You mentioned something of great interest in a previous post. I think you called it a "couple". But today you mentioned something that I have been thinking about for that last three weeks. It was Degrees Of Freedom.
Well of course I disagree. You might have noticed that I have alternate interpretations. They may lead to the same result in the end. But the path I took is different. I see different conclusions along the way. The degrees of freedom that you talk about look like directions in time to me. I believe that would be something new. A more basic understanding of time and how it works. Of course if you keep instructing me on how to actually apply the math to this. Mainstream science might take notice. You know the math is just a fringe benefit to me. I'll probably have a time machine up and running before I fully develop the math for it. I see six degrees of freedom for torque. Two directions in each orthogonal plane of 3-D space. So you might be either referring to a plane or a line as a degree of freedom if I try to interpret what you say from context. I merely chose a direction as being a direction in time. One of those six directions is responcible for the forward flow of time. With the gyroscope it is only possible to access four out of the six. And only three at a time. The open ended orthogonal torque would occur in orthogonal time. Any transfer of force in orthogonal time would appear to be instantaneous. And that is what my current experiment is all about. You might like it. I'm coupling two gyroscopes together in an attempt to create two spiral force directions that cause the device to move in a more stabilized pattern. The double spiral could be described as a double helix. If it works as planned, then I might try building a device that accesses another direction of time.
As for the math, you caught my interest when you mentioned a "couple". I think that might be a mathematical area that would cover some of my current experimentation. But there appears to be more than one way to couple torques together.
This last month I got a mini mill from Harbor Frieght. This thing cuts steel like butter. I just love to fabricate stuff.