How Does Applying Forces to a 3D Axis System Affect Its Motion and Stability?

[Jim Lundquist]

Please help me solve a problem that has puzzled me for years. I am not a mathematician or a physicist, so please bear with me. Imagine the apparatus pictured to be a depiction of the x, y and z axes. The x is represented by the green stick. The y is represented by the yellow stick, and the z by the blue stick. Let’s place an arbitrary range of each axis as -10 to +10. At -10 and +10 of each axis is a potential force represented by the arrows. When applied, it will be a constant force, such that when initiated in one axis only to the points at -10 and +10 that axis would scribe a circle. Now imagine a hypothetical sphere “enclosing” the apparatus whose x, y and z axes would correspond exactly to the x, y and z axes of the stationary apparatus. The imaginary sphere will never move, and will serve as a reference or standard. So the center of the imaginary sphere is x=0, y=0 and z=0…the same co-ordinates of the center point of the stationary apparatus. Now, apply an equal and constant force to each axis in the direction of their arrows at exactly the same moment. With no “knowledge” of the other axes, anyone axis will always “think” that it is fulfilling its mission of scribing a circle. Please forgive the anthropomorphism.

Questions:

Will the center point of the apparatus remain at the center point of the imaginary sphere or be displaced in a particular direction?

Since the end points of each axis will no longer scribe a circle, what will be the trajectory of each end point?

Will the end points of each axis scribe identical paths?

If this force could be applied indefinitely, would the end points eventually scribe a sphere?

If you reverse the direction of the forces on just one axis, how will it affect the behavior of the apparatus? Thank you!

 

[andrewkirk]

Questions:

Will the center point of the apparatus remain at the center point of the imaginary sphere or be displaced in a particular direction?

Since the end points of each axis will no longer scribe a circle, what will be the trajectory of each end point?

Will the end points of each axis scribe identical paths?

If this force could be applied indefinitely, would the end points eventually scribe a sphere?

If you reverse the direction of the forces on just one axis, how will it affect the behavior of the apparatus?

Thank you!

Surprisingly, the problem turns out to be much easier than it looks.

As described, a rotation is imparted to the rigid object in each of the three indicated directions.

The thing that simplifies all this is that rotations can be represented as vectors, pointing along the axis of rotation, and rotations add just as vectors do!

The rotations applied to the yellow, green and blue rods are represented by vectors, all the same magnitude, pointing in the green (away from us), blue (towards us) and yellow (down) directions.

The sum of those vectors is a vector pointing to the right and about 35 degrees below the plane of the blue and green rods. The magnitude of the vector is the length of the internal diagonal of a cube whose edges are the lengths of the individual vectors, which is $\sqrt 3$ times those lengths.

This means that the whole assembly will rotate around an axis aligned in the direction I described, at a rate that is $\sqrt3$ times each of the individual rotation rates. The central point of the assembly ('the origin') does not move.

In 3D space the combined rotation vector has its tail at the origin (0,0,0) and its head at the point (1,-1,1).

If one of the component rotations is reversed, the sign of the respective component of the vector (1,-1,1) is reversed, so that the vector retains its length but points to a corresponding point in a different octant (just as in 2D there are four quadrants in the number plane, in 3D there are eight octants).

The six end points will describe two circles around the axis of rotation, three of them (the two leftmost ones and the top one as we look at the picture) describing a circle along a line of Latitude in the Southern Hemisphere of the sphere (like a Tropic of Capricorn). and the other three following the Tropic of Cancer. Here I sue North and South as determined by the direction of rotation, using the same convention as is used with the Earth, not by the up and down directions in the photo.

 

[Jim Lundquist]

Thank you very much for taking the effort to explain this to me. If I am understanding you correctly, the tails (center) of the assembly will always remain at (0,0,0) and the motion of the heads or end points on all three axes, viewed heads toward tails, would appear to move as follows:

The entire assembly would be in a perpetual wobble under a constant and equal force applied on each axis as previously described. Like I said before, I am not a mathematician or physicist. My degrees are in psychology and biology and have just a basic understanding of vector math, but it would be interesting to see an animation of this model and the effect of varying the magnitude of the force applied on each axis. I imagine at zero gravity in a vacuum the wobble would continue even if the forces on all three axes were simultaneously removed?

 

[andrewkirk]

Yes, that's right.

It won't wobble though. It'll just rotate smoothly around the diagonal axis described. Changing the torques (it's better to think of them as torques than forces, 'torque' being the rotational equivalent of a force - it's a twist) will change speed of rotation and the direction of the axis of rotation but, at any point in time, the sphere will be rotating around a single axis (which will not be aligned with any of the three shown axes).

The (rather deep and beautiful) maths underlying this is that the set of rotations in 3D form a group under the action of combining them together. It's like Rubik's cube.

 

[A.T.]

The entire assembly would be in a perpetual wobble under a constant and equal force applied on each axis as previously described.

Not sure what you mean by "wobble". When it's already spinning, then under a torque that is not aligned with the angular velocity, you will have precession:



When the body had different moments of inertia around the three axes and would float freely, you could also see this:

https://en.wikipedia.org/wiki/Tennis_racket_theorem

 

[Jim Lundquist]

Thank you again for your explanations and patience, but I have been working on this problem in “mindspace” for several decades when the answer actually lies in a field, mathematics, that is relatively alien to me. In other words, I tend to try to picture the model in motion. Being 70 years old rather limits me in a pursuit to achieve any real mathematical skill. I had originally thought that my model would eventually scribe a sphere after some number of rotations following an unknown path. Please allow me to approach the problem from reverse. Knowing that a point is not a thing but a place, can we imagine, for the sake of discussion, that it is a thing the size of the period (.) on this page. If the axes in the model are lines composed of a set of these collinear points, and the surface of a hollow sphere is also composed of such points, is there such a rotation of these axes within this hollow sphere that will eventually scribe this sphere such that the rotation is fluid (most efficient, or the forces/torques are not artificially manipulated) and before the axes return to their original orientation? This may require that the “head” point of each axis must cross each point on the hollow sphere…but I don’t know. If this question is nonsensical in your highly structured discipline, please indulge this lay person one more time.

 

[A.T.]

A sphericaly symetrical object with no external forces after the initial spin up, will spin around a constant axis and reach the initial orientation after one revolution.

 

[Jim Lundquist]

Not sure what you mean by "wobble". When it's already spinning, then under a torque that is not aligned with the angular velocity, you will have precession:



When the body had different moments of inertia around the three axes and would float freely, you could also see this:

https://en.wikipedia.org/wiki/Tennis_racket_theorem

The tennis racket theorem (which I tried, and works as described) discusses first and third principal axes and a second principal axis also called an intermediate axis. If my model illustrates equal forces (torques) applied simultaneously in three dimensions, which of these axes would be considered first and third and intermediate? Each axis in my model describes its own plane, if rotated independently. The tennis racket is being tossed with a planar bias (the plane described by the axis drawn from head to handle), while my model has no such bias.

 

[A.T.]

The tennis racket theorem (which I tried, and works as described) discusses first and third principal axes and a second principal axis also called an intermediate axis. If my model illustrates equal forces (torques) applied simultaneously in three dimensions, which of these axes would be considered first and third and intermediate? Each axis in my model describes its own plane, if rotated independently. The tennis racket is being tossed with a planar bias (the plane described by the axis drawn from head to handle), while my model has no such bias.

If all your axes have the same moment of inertia, then there is no such distinction.

 

[Jim Lundquist]

A sphericaly symetrical object with no external forces after the initial spin up, will spin around a constant axis and reach the initial orientation after one revolution.

Then I guess you are saying that there is no possible "natural" rotation of the x, y and z axes that could scribe a sphere with their "head" points...even if the forces applied on each axis were "tweaked" as variables in some formula?

 

[A.T.]

Then I guess you are saying that there is no possible "natural" rotation of the x, y and z axes that could scribe a sphere with their "head" points...even if the forces applied on each axis were "tweaked" as variables in some formula?

I was talking about no forces, after the initial spin up. If you keep steering then it can reach any orientation.

 

[Jim Lundquist]

I was talking about no forces, after the initial spin up. If you keep steering then it can reach any orientation.

No...steering would not be allowed. I was referring to the original forces not being equal as originally stated. Instead of the force applied on each axis being, for example 4N, 4N, 4N, it would be 5N, 5N, 2N. Last question...lol.

 

[A.T.]

Ok then see post#7.

 

[Jim Lundquist]

Ok then see post#7.

I'm sorry, but I may be the densest thing going. In the model, the arrows represent constant force or torque, not an instantaneous pulse. Imagine them to be rocket engines all producing equal and constant force in the direction of the arrows. At t=0, the blue or x-axis appears in the model to be nicely horizontal, but after rotation begins, it is no longer in a horizontal attitude, but one that has moved into the original y and z planes. If the model were animated, it would show a different orientation at one second into the animation than it would at zero and at two seconds. Wouldn't the new common axis, as described in post #2 be different in each time frame? If an angle of axis can be calculated from a motionless model at t=0, then it seems that it would have to be recalculated at t=1 as the original orientation of the axes has changed relative to the imaginary control sphere. I'm sorry. I don't mean to be argumentative. I'm just trying to wrap my head around this. It seems like in this day of animation of various physical phenomena, there should be one to illustrate this one. The old gyroscope thing just does not cut it.

 

[andrewkirk]

In the model, the arrows represent constant force or torque, not an instantaneous pulse. Imagine them to be rocket engines all producing equal and constant force in the direction of the arrows.

As long as the torques are constant, they combine to make a single constant torque around the axis described in post 2. Starting from rest, the rigid assembly rotates at an ever-increasing rotational rate around that axis of rotation. The axis of rotation never changes because the net torque is aligned with the direction of rotation. If not, precession would occur, as AT pointed out.

Another way to try visualising it is as follows: Imagine inserting another rod - red, to distinguish it from the others - into the joining blob at the centre. The rod sticks out rightwards and down, equidistant between the right-hand ends (as seen in the picture) of the blue and green rods and the bottom end of the yellow rod.

Now start twirling that red rod anticlockwise (as seen from its end), while keeping the blob in the middle of the assembly still. The nearest ends of the other three rods will twirl around your hand that is holding the red one. If you could set up a mechanism that constantly increases the rotation rate, you will have the same effect as with constant forces applied at the other six points.

You are right, an animation would make this much easier to see, but I don't know how to make them.

 

[A.T.]

In the model, the arrows represent constant force or torque, not an instantaneous pulse. Imagine them to be rocket engines all producing equal and constant force in the direction of the arrows.

So the net torque is fixed in the frame of the object. Since you start from rest, the spin axis will remain fixed in the initial non-rotating frame.

 

[Jim Lundquist]

As long as the torques are constant, they combine to make a single constant torque around the axis described in post 2. Starting from rest, the rigid assembly rotates at an ever-increasing rotational rate around that axis of rotation. The axis of rotation never changes because the net torque is aligned with the direction of rotation. If not, precession would occur, as AT pointed out.

Another way to try visualising it is as follows: Imagine inserting another rod - red, to distinguish it from the others - into the joining blob at the centre. The rod sticks out rightwards and down, equidistant between the right-hand ends (as seen in the picture) of the blue and green rods and the bottom end of the yellow rod.

Now start twirling that red rod anticlockwise (as seen from its end), while keeping the blob in the middle of the assembly still. The nearest ends of the other three rods will twirl around your hand that is holding the red one. If you could set up a mechanism that constantly increases the rotation rate, you will have the same effect as with constant forces applied at the other six points.

You are right, an animation would make this much easier to see, but I don't know how to make them.

Thank you...It is starting to make sense to me now, but wouldn't the relative angle of the red rod be at 45 degrees to the blue, green and yellow rods instead of 35 degrees as stated in post #2? I am creating a new model to illustrate your suggestion. It appears that the rotation of the model would scribe to opposing cones with (0,0,0) at the apex/vertex of each. BTW, this model was meant to be a simplified illustration of my avatar. I want to show the mechanics of the early universe that consisted of nothing but a unified force field composed of absolutely dense linear force focused at a center point and the transformation of this force vector arrangement to rotational (torque) mechanics. Since this pre-Big Bang universe had no up, down, left or right, imagine each line in the avatar being part of a triad of x, y and z axes all with their own corresponding arrows of motion, as in my simplified model. This rearrangement of linear force made 3D matter and energy possible while still manifesting the original force in the form of gravity. I know this site regards this as metaphysics, but all science has its roots in philosophy. I promise not to bring it up again.

 

[andrewkirk]

wouldn't the relative angle of the red rod be at 45 degrees to the blue, green and yellow rods instead of 35 degrees as stated in post #2?

That's what I thought at first too. The reason it's not is as follows: The diagonal is the line segment that cuts right through the centre of a cube between opposite corners. In a 3D coordinate system this is the line segment from the point (0,0,0) to the point (1,1,1). To find the nature of this line we can proceed as follows:

First consider the line segment from (0,0,0) to (1,1,0), which runs diagonally along the base of the cube. This is the hypotenuse of a 45 degree right-angled triangle, and has length $\sqrt 2$, by Pythagoras' Theorem. That's where the 45 degrees, that we expected, comes in.

Now the red rod is the hypotenuse of the right-angled triangle whose base is the segment (0,0,0)-(1,1,0) and whose vertical side is the segment (1,1,0)-(1,1,1). This is a right-angled triangle with adjacent and opposite sides of length $\sqrt 2$ and $1$. Hence by Pythagoras the hypotenuse has length $\sqrt{(\sqrt 2)^2+1^2}=\sqrt 3$ and whose angle between the horizontal line (0,0,0)-(1,1,0) and the hypotenuse (0,0,0)-(1,1,1) has tangent $1/\sqrt 2$. So that angle is $\tan^{-1}\left(1/\sqrt 2\right)=35.3^\circ$.

Note that this is not the angle to any of the original three rods. It is the angle to the horizontal plane. The angle to each of the rods is the 90 degrees minus that, which is about 55 degrees. To see that, note that the line (1,1,0)-(1,1,1) is the adjacent side of the triangle and is also parallel to the vertical axis. The angle of the red rod to that adjacent side is 90 deg minus the 35 deg angle to the plane. By the parallelism we just noted, that must also be the angle the red rod makes with the vertical axis. Since the problem is symmetrical between the three original axes, we conclude that that must also be the angle the red rod makes with each of the other two axes.

 

[Jim Lundquist]

Now you tell me...lol. I had already built the new model using 45 degrees, and as you predicted, it returned to the original orientation after one rotation...with no "wobble". My intuition (but when has that ever been right) tells me that 35 degrees is not going to be symmetrical and will create a wobble, but we'll see.

 

[Jim Lundquist]

That's what I thought at first too. The reason it's not is as follows: The diagonal is the line segment that cuts right through the centre of a cube between opposite corners. In a 3D coordinate system this is the line segment from the point (0,0,0) to the point (1,1,1). To find the nature of this line we can proceed as follows:

First consider the line segment from (0,0,0) to (1,1,0), which runs diagonally along the base of the cube. This is the hypotenuse of a 45 degree right-angled triangle, and has length $\sqrt 2$, by Pythagoras' Theorem. That's where the 45 degrees, that we expected, comes in.

Now the red rod is the hypotenuse of the right-angled triangle whose base is the segment (0,0,0)-(1,1,0) and whose vertical side is the segment (1,1,0)-(1,1,1). This is a right-angled triangle with adjacent and opposite sides of length $\sqrt 2$ and $1$. Hence by Pythagoras the hypotenuse has length $\sqrt{(\sqrt 2)^2+1^2}=\sqrt 3$ and whose angle between the horizontal line (0,0,0)-(1,1,0) and the hypotenuse (0,0,0)-(1,1,1) has tangent $1/\sqrt 2$. So that angle is $\tan^{-1}\left(1/\sqrt 2\right)=35.3^\circ$.

Note that this is not the angle to any of the original three rods. It is the angle to the horizontal plane. The angle to each of the rods is the 90 degrees minus that, which is about 55 degrees. To see that, note that the line (1,1,0)-(1,1,1) is the adjacent side of the triangle and is also parallel to the vertical axis. The angle of the red rod to that adjacent side is 90 deg minus the 35 deg angle to the plane. By the parallelism we just noted, that must also be the angle the red rod makes with the vertical axis. Since the problem is symmetrical between the three original axes, we conclude that that must also be the angle the red rod makes with each of the other two axes.

Let me see if this illustrates what you are saying. If I insert the red rod into the center "blob", as you describe, and make that rod a vertical axis, and if the line segments of the +x, -x, +y, -y, +z, -z are of equal length, and if I make the ends of each segment a pencil, I imagine that the "head" of the +x, -y and -z would be resting on a flat sheet of graph paper with the red rod penetrating below the surface of the paper (for convenience). When I spin the red rod, the +x, -y and -z "heads" or "pencils" would scribe a circle on the graph paper with the red rod being the center of the circle. This would not be an ellipse, but a circle, and the "pencils" would never leave the surface of the graph paper...correct? Now if I put a "roof" on this model of another piece of graph paper, I imagine that the -x, +y and +z "pencils" would also be in contact with the graph paper. When I spin the red rod, I imagine that the -x, +y and +z pencils would also scribe a circle. If I remove both pieces of graph paper and overlay one over the other, the circles scribed on each piece of graph paper would also overlay...correct? If we don't illustrate anything else in this exercise, we will illustrate that a picture is worth a thousand words...lol. It may also illustrate the different tools individuals use in the thought process, for example words, mathematics and picture visualization.

 

[andrewkirk]

Yes that sounds correct.

 

[Jim Lundquist]

Yes that sounds correct.

Thank you, Andrew. I really appreciate all of your help.

 

[Jim Lundquist]

After pondering this discussion, I am still left with this one question: If each axis examined separately is initially attempting to rotate in its own described plane and counter to the other planes, why would there be a bias to establish the “new” rotational axis…for example why is an axis established including the set of points ((-1, -1, 1), (0, 0, 0), (1, 1, -1)) as opposed to ((-1, -1, -1), (0, 0, 0), (1, 1, 1)) OR ((1,- 1, -1), (0, 0, 0), (-1, 1, 1)) OR ((1, 1, -1), (0, 0, 0), (-1, -1, -1))? It seems like there would have to be a variable to describe this “new” axis in a model that is described as having no variables…just the constant and even application of force (torque) in 3 orthogonal dimensions. Are you assuming an observational frame of reference? Thank you.

 

[andrewkirk]

I'm not sure I understand your question, but it sounds like you are troubled by the fact that the rotation is around an axis that no individual force is creating a pure torque around.

This can seem unintuitive, but it's just vector addition. If a plane is facing North and flying North relative to the air, and the air is moving East (ie the wind is from the West), the plane will move in a direction between North and East, even though neither the plane's engines nor the wind are pushing purely in that direction.

Another example is that if you want to swim directly across a river you need to head slightly diagonally upstream, so that the upstream component of your motion relative to the water offsets the downstream drift from the current. Your motion relative to the river bed is directly across the river, even though neither the force the water applies to you (in reaction against your strokes) nor the direction of the current points in that direction.

In this problem, the various torques add together as vectors to create a net torque that is around the final axis, even though none of the individual torques is purely around that axis. They each have components that are around that axis though.

 

[A.T.]

a model that is described as having no variables…just the constant and even application of force (torque) in 3 orthogonal dimensions.

What do you mean by "no variables"? The torques you apply are the variables, which determine the rotation.

 

[Jim Lundquist]

A.T., torques are equal on all three axes and continuous, as if the arrows shown in the OP were rocket engines.

Andrew, if there are 8 possible directional combinations of force (torque) applied in the model, as shown in the chart below, and the “inverted” combinations cancel out the “standard” combinations, then there should be 4 possible axes of rotation. Is this finally correct? If so, I have one remaining question: Is there a formula that could be applied to the original model such that the torque values could be tweaked (no longer be equal) so that (0,0,0) in the model remains at (0,0,0) of a fixed imaginary reference sphere? We stated earlier that the end points of the x, y and z axes rotating around the new resultant axis would scribe a circle (at the end points) and a cone (including the line segments of each axis) with (0,0,0) being the apex of the cone. Basically, I want to know if the model could be spun in such a way that the end points would eventually scribe a sphere.

 

[A.T.]

A.T., torques are equal on all three axes and continuous, as if the arrows shown in the OP were rocket engines.

Yes, that gives you a total torque along one of the unit cube diagonals.

Basically, I want to know if the model could be spun in such a way that the end points would eventually scribe a sphere.

Basically, you are asking the same question over and over again.

 

[Jim Lundquist]

Thank you...so the answer is no, a sphere cannot be mathematically described by any kind of 3D rotation. A sphere can only be described by a 2D circle rotating around a 1D diameter.

 

[andrewkirk]

a sphere cannot be mathematically described by any kind of 3D rotation

That is true for constant-torque rotations, which is all that has been considered here. If the torque were varied over time, the axis of rotation could be made to vary and it is conceivable that the scheme by which the torque is varied over time could be designed such that every point on the sphere must be traversed by one of the rockets at some time.

I imagine it would be difficult to try to design such a scheme though. Nor am I convinced that it is possible, because then the trajectories of the rockets would have to be space-filling curves. If we weakened the requirement to one that every point on the sphere must at some point have a rocket pass within distance $\epsilon$ of it, where we can specify $\epsilon$ to be as small as we like, then it should be easy enough to design such a scheme.

It's easy enough to colour in a square using a pencil with point-width 1mm, but much more difficult if the point-width is zero.

 

[Jim Lundquist]

Thank you, Andrew. I couldn't have asked for a better explanation. It is precisely what I was looking for. It is interesting when you say "It's easy enough to colour in a square using a pencil with point-width 1mm, but much more difficult if the point-width is zero." Can the "width" of a line be smaller than Planck Distance? If not, then maybe if given "infinite" time it is possible. Thanks again for your help. It has been greatly appreciated!