# I Problem in 3D Rotation

1. May 7, 2017

### 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, any one 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!

2. May 7, 2017

### andrewkirk

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.

3. May 7, 2017

### 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?

Last edited: May 7, 2017
4. May 8, 2017

### 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.

5. May 8, 2017

### A.T.

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

Last edited: May 8, 2017
6. May 8, 2017

7. May 8, 2017

### 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.

8. May 8, 2017

### Jim Lundquist

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.

9. May 8, 2017

### A.T.

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

10. May 8, 2017

### Jim Lundquist

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?

11. May 8, 2017

### A.T.

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

12. May 8, 2017

### Jim Lundquist

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.

13. May 8, 2017

### A.T.

Ok then see post#7.

14. May 8, 2017

### Jim Lundquist

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.

15. May 8, 2017

### andrewkirk

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.

16. May 9, 2017

### A.T.

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.

17. May 9, 2017

### Jim Lundquist

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.

18. May 9, 2017

### andrewkirk

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.

19. May 9, 2017

### 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.

20. May 10, 2017

### Jim Lundquist

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.