Thread: Gravitational Potential of a Ring Mass in the Plane of the Ring

1. Established Member
Join Date
Feb 2012
Posts
451

Gravitational Potential of a Ring Mass in the Plane of the Ring

Clearly the outline of solution for the force due to a ring distribution along the axis can be found in any elementary text on electromagnetism.

In particular, I am interested in the self attraction of the ring. From any point on the ring, all of the mass of the ring is towards the center of the ring. This should provide a central acceleration on the ring. I am wanting to solve for the rotation rate needed to balance this acceleration.

For simplicity, let's assume we are using units where G=1, the radius of the ring is 1 and the mass of the ring is 2pi. This gives a linear density, l, of 1. By symmetry the force must be central.

Let us consider the force felt by a point at an angle of 0 due to the mass at an angle of theta. We also need the distance to that point. I get:

Integrating this over the ring gives us an infinite value. Infinite is much larger than the finite values of a central mass, yet stable rings are found in nature.

Where is my mistake?

2. Established Member
Join Date
Jun 2009
Posts
1,044
I'm not quite following you. I get stuck when you introduce the distance. It might be easier to sketch it.

I would personally appeal to the symmetry of the problem; just as you would to solve for force along the central axis. Choose a point, P, on the ring. Opposite to that point, there is a complementary point, call it Q. For each point P that you choose, there is one and only one Q. By symmetry, as you said, the force points to the very center of the ring. Now, since we already know the force points toward the center, we only need to calculate that component of the force. This will make you integrate theta from 0 to Pi(and then take the answer times 2), instead of 0 to 2Pi.

Now imagine a circle. Start at the point (-1, 0). You're at the far left of the circle, along the x-axis. The center of that circle points directly to the 'right' or the x-axis unit vector's direction. The force felt only needs to include the x component. The distance is the full distance, but only Force*cos(theta) matters. Just thinking about it, I would first try to integrate it in polar coordinates, suspecting the distance will come out as simply cos(theta) instead of sqrt(x2 + y2). In which case the limits of integration might even be 0 to Pi/2. You are starting from a point on the ring after all.

edit: ...your distance is polar to begin with! I gotta stop posting after work. With the circle I setup above, distance=2cos(theta). Technically, you'd put yourself at the origin but then the circle is centered on (1, 0). You then integrate for cos between 0 and Pi/2. Take the integral as double, being an even function integrated over half range. Comes out to the same thing as taking theta/2.

sin(theta)?
Last edited by ShinAce; 2012-Aug-28 at 11:52 PM.

3. Originally Posted by utesfan100
yet stable rings are found in nature.
Stable self-gravitating rings are found in nature? Could you tell us more about these rings, please?

4. Member
Join Date
Aug 2012
Posts
53
Can't find a mistake.

If you think about it, this is not really a natural situation at all, since you have to have infinite volume density to get finite surface and line densities. As we know, the case of finite surface density on a spherical shell turns out to be mathematically well-behaved in spite of that (same gravitational field on the shell as if its total mass were a central point mass). For a ring, though, the fraction of the total mass which is directly adjacent to a given point along the circumference is relatively greater than it is for a spherical shell (infinitesimal versus infinitesimal^2, if you see what I mean). Thus, it doesn't seem implausible that that fraction could produce a diverging central acceleration, as you expression indicates.

If you still don't trust your maths, just model it numerically. If the system were well-behaved, the result should converge as you decrease delta_theta. If it doesn't, there you are.

ETA: Yep, a quick and very dirty numerical approach supports your result. For a spherical shell, I get the following:

 1 0.5 0.25 0.125 0.0625 0.03125 28.6515 14.3244 7.16203 3.58099 1.79049 0.89525

The top row gives the angular size (as measured from the centre, in degrees) of the "vicinity" of a point. The bottom row gives an arbitrarily but consistently scaled numerical approximation of the central acceleration due to the mass within this vicinity. As the size approaches zero, do does its contribution, which suggests that a summation over the whole sphere would have a finite result. For a circular ring, though, it changes to this:

 1 0.5 0.25 0.125 0.0625 0.03125 28.6515 28.6488 28.6481 28.648 28.6479 28.6479

Now, as the size approaches zero, its contributions approach a constant greater than zero. Whenever infinitesimal regions give you finite contributions, you're pretty much screwed, I'm afraid.
Last edited by onomatomanic; 2012-Aug-29 at 12:56 AM.

5. Established Member
Join Date
Feb 2012
Posts
451
Originally Posted by ShinAce
I'm not quite following you. I get stuck when you introduce the distance. It might be easier to sketch it.

I would personally appeal to the symmetry of the problem; just as you would to solve for force along the central axis. Choose a point, P, on the ring. Opposite to that point, there is a complementary point, call it Q. For each point P that you choose, there is one and only one Q. By symmetry, as you said, the force points to the very center of the ring. Now, since we already know the force points toward the center, we only need to calculate that component of the force. This will make you integrate theta from 0 to Pi(and then take the answer times 2), instead of 0 to 2Pi.

Now imagine a circle. Start at the point (-1, 0). You're at the far left of the circle, along the x-axis. The center of that circle points directly to the 'right' or the x-axis unit vector's direction. The force felt only needs to include the x component. The distance is the full distance, but only Force*cos(theta) matters. Just thinking about it, I would first try to integrate it in polar coordinates, suspecting the distance will come out as simply cos(theta) instead of sqrt(x2 + y2). In which case the limits of integration might even be 0 to Pi/2. You are starting from a point on the ring after all.

edit: ...your distance is polar to begin with! I gotta stop posting after work. With the circle I setup above, distance=2cos(theta). Technically, you'd put yourself at the origin but then the circle is centered on (1, 0). You then integrate for cos between 0 and Pi/2. Take the integral as double, being an even function integrated over half range. Comes out to the same thing as taking theta/2.

sin(theta)?
I suppose I should explain my symbols more precisely.

Let us have a ring of mass 2piM and radius R. This gives us l=M/R. Let us put the origin at the center, and place the ring in the XY plane. Let us now consider the force on the ring at the point where the ring meets the negative Y axis. Let us also define theta to be the angle from the center to the differential region on the ring being considered.

By bisecting the radii connecting these two points, we bisect the equilateral triangle formed by the two radii and the segment connecting the two points. This allows us to show that .

Drawing a perpendicular from the differential region being considered to the radius to the point we are examining the force at creates a triangle sharing the angle at the point we are considering with one of the two half isosceles triangles. This means the two triangles are equivalent, which means the angle at the differential region is theta/2. The proportion of the force towards the center is then

This gives us the following for the force:

This integral increases without bound. onomatomanic showed numerically that this is reasonable.

6. Established Member
Join Date
Feb 2012
Posts
451
Originally Posted by StupendousMan
Stable self-gravitating rings are found in nature? Could you tell us more about these rings, please?
Well, if the self force on a ring is infinite, then the self force of any particular ring of Saturn should greatly exceed the acceleration caused by Saturn.

7. Established Member
Join Date
Feb 2012
Posts
451
Originally Posted by onomatomanic
Can't find a mistake.

If you think about it, this is not really a natural situation at all, since you have to have infinite volume density to get finite surface and line densities. As we know, the case of finite surface density on a spherical shell turns out to be mathematically well-behaved in spite of that (same gravitational field on the shell as if its total mass were a central point mass). For a ring, though, the fraction of the total mass which is directly adjacent to a given point along the circumference is relatively greater than it is for a spherical shell (infinitesimal versus infinitesimal^2, if you see what I mean). Thus, it doesn't seem implausible that that fraction could produce a diverging central acceleration, as you expression indicates.

If you still don't trust your maths, just model it numerically. If the system were well-behaved, the result should converge as you decrease delta_theta. If it doesn't, there you are.

...
Now, as the size approaches zero, its contributions approach a constant greater than zero. Whenever infinitesimal regions give you finite contributions, you're pretty much screwed, I'm afraid.
This seems to imply that the binding energy for this configuration is infinite.

Since this is mathematically parallel to the case of a ring charge, does this mean that a uniform ring of charge has infinite binding energy?

8. Order of Kilopi
Join Date
Jan 2010
Posts
3,586
Have you tried letting the ring be a torus of some finite inner radius r and then taking the limit where you let r go to zero? I'm not saying it would work, i haven't tried, it's just an idea. You can sometimes get rid of infinities that way.

9. Member
Join Date
Aug 2012
Posts
53
Originally Posted by utesfan100
This seems to imply that the binding energy for this configuration is infinite.
Not necessarily, you'd have to do the maths rigorously. Energy is the path integral of the force, and the force only diverges for points on the circumference itself, while it's certain to be finite for all points a finite distance off the circumference. Thus, one integrates across "competing infinities" - an infinitesimally narrow region of infinite value. Such a contribution can turn out to be infinite, or finite, or even zero, there's no straightforward way of telling without actually doing the work, as far as I know.

Originally Posted by utesfan100
Since this is mathematically parallel to the case of a ring charge, does this mean that a uniform ring of charge has infinite binding energy?
The inference sounds plausible to me, considering that the only difference between the two situations is that like masses attract while like charges repel, which should have no more effect than to flip the overall result from positive to negative. If this turned out to be the case, would you find that any more counterintuitive than the infinite self-energy of any point charge, in the classical treatment?

10. Established Member
Join Date
Feb 2012
Posts
451
Originally Posted by caveman1917
Have you tried letting the ring be a torus of some finite inner radius r and then taking the limit where you let r go to zero? I'm not saying it would work, i haven't tried, it's just an idea. You can sometimes get rid of infinities that way.
Yes. I quickly run into elliptic integrals for which my intuition on infinities fail.

11. Established Member
Join Date
Feb 2012
Posts
451
Originally Posted by onomatomanic
Would you find that any more counterintuitive than the infinite self-energy of any point charge, in the classical treatment?
I suppose the idea that compressing a density to one less dimension leads to resolvable self energy, but compressing two or more dimensions leads to infinite binding energy, requiring the physical thickness or discrete nature of the medium (at with Saturn's rings) to be considered.

It would be interesting to try to demonstrate this principle using Gauss's Law in an arbitrary dimension.

ETA:

Observing that 3/4 of the mass of the Milky Way is more disk-like than spherical, It might be interesting to calculate the impact on the inverse square force this causes.
Last edited by utesfan100; 2012-Aug-29 at 04:55 AM. Reason: Thought

12. Member
Join Date
Aug 2012
Posts
53
Originally Posted by utesfan100
[...] compressing a density to one less dimension leads to resolvable self energy, but compressing two or more dimensions leads to infinite binding energy [...]
The much-generalized form of that realization is what is referred to as the "privileged character of 3+1 spacetime", I believe:

Originally Posted by utesfan100
[...] requiring the physical thickness or discrete nature of the medium (at with Saturn's rings) to be considered.

[...]

Observing that 3/4 of the mass of the Milky Way is more disk-like than spherical, It might be interesting to calculate the impact on the inverse square force this causes.
The more useful analogy for both of those cases may be the spherical shell, though - squash it flat and replace the central section with a blob. The approach will involve surface densities and a directly adjacent region of infinitesimal^2 size. So I don't imagine you'll run into problems there, even if you treat them as having zero thickness.

Originally Posted by utesfan100
It would be interesting to try to demonstrate this principle using Gauss's Law in an arbitrary dimension.
I thought about that, but if you actually change the dimensionality of the space and not just that of the object, the distance-dependence of force changes as well. I quite agree that such a demonstration would be interesting, but it may end up not tell you much about the case you're considering.

13. Originally Posted by utesfan100
Well, if the self force on a ring is infinite, then the self force of any particular ring of Saturn should greatly exceed the acceleration caused by Saturn.
Bad example. The rings of Saturn are not self-gravitating, since the gravitational force on any ring particle from Saturn is far, far, larger than the gravitational force from the other ring particles. In the case of Saturn's rings, the gravitational effects of the ring particles on each other are simply (heh) perturbations to the dominant force from the planet.

14. Established Member
Join Date
Feb 2012
Posts
451
Originally Posted by StupendousMan
Bad example. The rings of Saturn are not self-gravitating, since the gravitational force on any ring particle from Saturn is far, far, larger than the gravitational force from the other ring particles. In the case of Saturn's rings, the gravitational effects of the ring particles on each other are simply (heh) perturbations to the dominant force from the planet.
As noted in post 11.

15. Established Member
Join Date
Feb 2012
Posts
451
Sketch of a proof, mathematicians will need more rigor.
Finite binding energy of D-1 surface
I think the proof stems from the idea that the field energy is proportional to the flux density in a region. If we have a mass distribution confined within some surface with finite field energy we can project it along the field to the bounding surface. The flux outside the surface will remain unchanged by this projection. Since the field energy of the original mass is finite and moving the mass outward to the bounding surface reduces the field energy, a surface density can produce a finite energy density. It must have a finite curvature at every point.

Infinite binding energy of D-N surface, N>1
If the mass is confined to a surface two dimensions less than our space we can examine the plain where the mass locally projects to a point. The flux density increases along this plane without bound as we approach the point, showing that the field energy must be infinite.

Originally Posted by onomatomanic
The much-generalized form of that realization is what is referred to as the "privileged character of 3+1 spacetime", I believe:

I thought about that, but if you actually change the dimensionality of the space and not just that of the object, the distance-dependence of force changes as well. I quite agree that such a demonstration would be interesting, but it may end up not tell you much about the case you're considering.
I believe that the graph you posted refers to the nature of the central force law. An inverse square law has unique stability properties.

The more useful analogy for both of those cases may be the spherical shell, though - squash it flat and replace the central section with a blob. The approach will involve surface densities and a directly adjacent region of infinitesimal^2 size. So I don't imagine you'll run into problems there, even if you treat them as having zero thickness.
If we squish it flat we get a disk, which will also have infinitesimal^2 size.

Any squishing of the spherical shell produces elliptic integrals. I would rather deal numerically than with elliptic integrals.

16. Member
Join Date
Aug 2012
Posts
53
Originally Posted by utesfan100
Infinite binding energy of D-N surface, N>1
If the mass is confined to a surface two dimensions less than our space we can examine the plain where the mass locally projects to a point. The flux density increases along this plane without bound as we approach the point, showing that the field energy must be infinite.
You've lost me. Could you expand that part a little bit, especially regarding your choice of the "plain" and "point" which yield this projection?

Originally Posted by utesfan100
Any squishing of the spherical shell produces elliptic integrals. I would rather deal numerically than with elliptic integrals.
Wise words.

17. Established Member
Join Date
Feb 2012
Posts
451
Originally Posted by onomatomanic
You've lost me. Could you expand that part a little bit, especially regarding your choice of the "plain" and "point" which yield this projection?
For the ring configuration we initially considered, locally the geometry is nearly a line, and D=3, N=2. If we select a plane perpendicular to the tangent line at a point and project an infinitesimal region about our point into this plane our mass will be a point. The projected flux density will increase without bound as we approach the point.

Similarly, if we have the shell of a torus in a 4D space (Say in the XYZ space, but flat in the w direction), we can select any point on the torus. This point will have an outward normal line in the XYZ space. This line and the w axis will define a normal plane, and the projection of an infinitesimal region about our test point on our torus to this plane will be a point.

In short, if we are embedding a D-N dimensional surface density within a D dimensional space, there will be an N dimensional tangent normal space at every point on the surface. A projection of an infinitesimal region about a point on our surface to this tangent space will reduce to a point. For N>1 the projected flux density will increase without bound. For N=1 the flux due to the surface density must be outward and normal to the surface.
Last edited by utesfan100; 2012-Aug-29 at 11:21 PM. Reason: Fix Tag

18. Member
Join Date
Aug 2012
Posts
53
^ Oh, okay, I think I get it now. The thing that makes the 1-dimensional subspace special is that there is no more distance-dependence of flux there, so if it's finite some distance from the surface, it must also be finite at the surface itself, yes? Impressively elegant!

I'm still not convinced that you can directly infer infinite field energy from locally infinite flux density, though. As a thought experiment, take the electric field between two capacitor plates of finite size and pinch the field lines closer and closer together halfway between the plates. In the limiting case, you get an infinite flux across an infinitesimally small region, just like for a section of the ring. If I'm not mistaken (and I'm not sure that I'm not, mind you), this transformation doesn't affect the energy content of the field as a whole, which both before and after the pinch must be equal to the energy spent in charging the capacitor in the first place.

To make that inference solid, you'd have to show that in our case, the volume integral over the region of infinite flux density actually yields infinite flux in spite of being infinitesimally small, i.e. that the divergence away from zero of the one "wins out" over the asymptotic approach of zero of the other.

19. Established Member
Join Date
Feb 2012
Posts
451
Originally Posted by onomatomanic
I'm still not convinced that you can directly infer infinite field energy from locally infinite flux density, though. As a thought experiment, take the electric field between two capacitor plates of finite size and pinch the field lines closer and closer together halfway between the plates. In the limiting case, you get an infinite flux across an infinitesimally small region, just like for a section of the ring. If I'm not mistaken (and I'm not sure that I'm not, mind you), this transformation doesn't affect the energy content of the field as a whole, which both before and after the pinch must be equal to the energy spent in charging the capacitor in the first place.
But what pinches the flux? The flux due only to the plates is normal to the two plates. A gradient in flux is also proportional to the electric field. The pinching of the flux must be due to some other field sources, and it must resemble a dipole at the center.

But a dipole is two point charges, with N=D, and we would expect an infinite potential near the dipole.

20. Established Member
Join Date
Jun 2009
Posts
1,044
Originally Posted by StupendousMan
Bad example. The rings of Saturn are not self-gravitating, since the gravitational force on any ring particle from Saturn is far, far, larger than the gravitational force from the other ring particles. In the case of Saturn's rings, the gravitational effects of the ring particles on each other are simply (heh) perturbations to the dominant force from the planet.
I didn't have time to read the paper linked in the footnotes, but this was interesting to find:
http://wjd.ita.uni-heidelberg.de/AST...2004-09-01.pdf

The author argues that as for the structure of Saturn's rings, the self(or proper) gravity is not negligible.

edit: self(proper)

21. Established Member
Join Date
Feb 2012
Posts
451
Originally Posted by ShinAce
I didn't have time to read the paper linked in the footnotes, but this was interesting to find:
http://wjd.ita.uni-heidelberg.de/AST...2004-09-01.pdf

The author argues that as for the structure of Saturn's rings, the self(or proper) gravity is not negligible.

edit: self(proper)
Very interesting paper. It actually argues that in the radial direction the self gravity is negligible, but in the vertical direction the local mass dominates.

Thus the particles will oscillate above and below the plane of the rings at a faster rate than their orbital period.

22. Order of Kilopi
Join Date
Feb 2005
Posts
7,182
A couple of things I was thinking about. Could a ring shaped space station be slowly assembled over deep time by injecting dust in a ring and then using field manipulation. Also--if bigger replacements for Large Hadron are made with the structure in a straight line--what effect might that have?

23. Order of Kilopi
Join Date
Sep 2004
Posts
5,446
Originally Posted by utesfan100
Clearly the outline of solution for the force due to a ring distribution along the axis can be found in any elementary text on electromagnetism.

In particular, I am interested in the self attraction of the ring. From any point on the ring, all of the mass of the ring is towards the center of the ring. This should provide a central acceleration on the ring. I am wanting to solve for the rotation rate needed to balance this acceleration.

For simplicity, let's assume we are using units where G=1, the radius of the ring is 1 and the mass of the ring is 2pi. This gives a linear density, l, of 1. By symmetry the force must be central.

Let us consider the force felt by a point at an angle of 0 due to the mass at an angle of theta. We also need the distance to that point. I get:

Integrating this over the ring gives us an infinite value. Infinite is much larger than the finite values of a central mass, yet stable rings are found in nature.

Where is my mistake?
Your integration is from -pi/2 to pi/2, and taking the limits of the improper integrals should come up with the right answer. I havent actually done the calculation yet, I will try to get back to it later.

24. Member
Join Date
Aug 2012
Posts
53
^ In the geometry laid out in the OP (with our test-mass at 0 pi), those integration limits represent the near semicircle (i.e. the semicircle centred on the test-mass). There are no obvious symmetries which would allow one to deduce the contribution of the far semicircle from the result, that I can see. Rather, the integration should be from 0 pi to 2 pi, or from 0 to pi using the obvious symmetry of the left-hand and right-hand semicircles. Ultimately, it doesn't really matter, because the integral of 1/sin(x) diverges both below and at zero.

25. Established Member
Join Date
Jan 2007
Posts
334
Originally Posted by onomatomanic
^ In the geometry laid out in the OP (with our test-mass at 0 pi), those integration limits represent the near semicircle (i.e. the semicircle centred on the test-mass). There are no obvious symmetries which would allow one to deduce the contribution of the far semicircle from the result, that I can see. Rather, the integration should be from 0 pi to 2 pi, or from 0 to pi using the obvious symmetry of the left-hand and right-hand semicircles. Ultimately, it doesn't really matter, because the integral of 1/sin(x) diverges both below and at zero.
Indeed , it doesn't matter : the expression int (dx/sinx) = ln ( tan(x/2)) gives both in the upper and lower limit a ln(0) result . Meaning infinite acceleration .

I think this result can be expected , because the system set up ( a 1 ( one!) dimensional ring having mass ) is not realistic . If the ring had a "gap" near the point of investigation one gets a non infinite result . Having mass just infinitively near the point of investigation gives infinity .
If one would consider however a ring existing between (r1, r2 ) and investigate at (r1+r2)/2 one might get a finite result , . I guess . Such a set up is more realistic
This model is hard te integrate however , I'm not sure . Haven't done it . But the denomimator of the integrand probably will not be infinte at the point of investigation .

The problem of infinity arises probably while one tends to compress matter in one dimension . A torus (3dim ) works , as a sphere works . Probably a disc (2dim ) works . A ring (1 dim ) doesn't work .

26. Established Member
Join Date
Feb 2012
Posts
451
Originally Posted by frankuitaalst
The problem of infinity arises probably while one tends to compress matter in one dimension . A torus (3dim ) works , as a sphere works . Probably a disc (2dim ) works . A ring (1 dim ) doesn't work .
Yes. I am convinced that a 1D ring is not solvable. For the reasons given in post #15 the binding energy will be infinite, and the derivative of this still infinite.

If we must simplify the 3D shape, a 2D surface is needed. The two that seem the closest to the geometry are the shell of a torus, or a thin annulus. When we make the radius of the torus or thickness of of the annulus smaller the acceleration will increase without bound.

An interesting question would be what shape surface would be normal to the flux at every point. This would likely be an egg shaped figure rotated about the Z axis.

27. Order of Kilopi
Join Date
Sep 2004
Posts
5,446
Originally Posted by utesfan100
Yes. I am convinced that a 1D ring is not solvable. For the reasons given in post #15 the binding energy will be infinite, and the derivative of this still infinite.

If we must simplify the 3D shape, a 2D surface is needed. The two that seem the closest to the geometry are the shell of a torus, or a thin annulus. When we make the radius of the torus or thickness of of the annulus smaller the acceleration will increase without bound.

An interesting question would be what shape surface would be normal to the flux at every point. This would likely be an egg shaped figure rotated about the Z axis.
The ring you presented in the OP is 2-d not one. for the setup you gave the dimensions are radial and azmuthal. The radial dimension just dosent show up in the integration. This is also a perfectly solvable (and fairly trivial) in cartesian coordinates. This means that there must be a solution in polar coordinates also.

Come to think of it, you cant use the standard force equation to find the self gravitation force on the ring. The derivitive of the potential energy equation has a discontinuity at that point, so the standard force equation dosent actually exist. That is why you are getting an infinite solution. I think it is Green's theorem to actually get the answer, but dont quote me on that. It is solvable tho, and it gives a finite solution, and it is the finite solution you would expect. It is just more math than I am willing to do here.

28. Originally Posted by korjik
It is solvable tho, and it gives a finite solution, and it is the finite solution you would expect.
JOOC, what is the solution that you would expect?

29. Member
Join Date
Aug 2012
Posts
53
Originally Posted by utesfan100
An interesting question would be what shape surface would be normal to the flux at every point. This would likely be an egg shaped figure rotated about the Z axis.
Good call, though the eggyness (ovoidity?) is less than I'd have expected:

This is the output of my, by now, less quick but still fairly dirty numerical model for a slice of the XZ plane centred on a point at which the ring intersects it, i.e. on (1,0). The red-blue lines show the force at each point (from red to blue, length of line proportional to strength of force), the green lines are normal to the former, and thus tangent to the equipotential surfaces. The Y-components aren't shown but are known to be zero from symmetry.

... I dunno, it seems quite self-explanatory to me, but that may not mean much. If more exposition is needed, let me know.

30. Order of Kilopi
Join Date
Sep 2004
Posts
5,446
Originally Posted by grapes
JOOC, what is the solution that you would expect?
In cartesian coords, F=-Gmrmp/r2x

Where
mr is the mass of the ring
mp is the mass of a little tiny bit of the ring
r is the radius of the ring
x is the x unit vector

with the ring centered on the origin and the point being measured at (1,0) on the xy plane

I think with the OP numbers it is F=-x, but I am too lazy to actually look

Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•