I finished, B is a lot harder, because it has few symmetry, also there was a typo in the formula in my previous post. Assume without loss of generality that A lies on the x-axis and B's y-coordinate is zero.
I cannot guarantee that the integrals have finite form primitives, perhaps someone could fill some values and evaluate them numerically to see the direction the vectors point?
EDIT: Does anyone want to know the math behind this? I think it's very annoying...
EDIT 2: I keep confusing R with r!!!!
In the last two integrals, r^2 (sin fi - 1) read r^2 sin fi - R^2. And in the third, r^3 (sin fi - 1) = r^2 ( r sin fi - R)
It might be easier to just fill up a toroidal volume with voxels of equal mass and then calculate the summed gravitational pull from all of them at various points -- roughly equivalent to numerical calculation of your triple integral, but easier for me to wrap my head around at any rate. :) You probably wouldn't need more than 10k voxels to get an adequate approximation.
Pyrel - an open-source rewrite of the Angband roguelike game in Python.
Assume without loss of generality that A lies on the x-axis and B's y-coordinate is zero.
I didn't quite understand. Wouldn't that mean that both A and B are on the x-axis? Perhaps you meant that A is on the x-axis and B on the y-axis?
Derakon wrote:
It might be easier to just fill up a toroidal volume with voxels of equal mass and then calculate the summed gravitational pull from all of them at various points -- roughly equivalent to numerical calculation of your triple integral, but easier for me to wrap my head around at any rate. :) You probably wouldn't need more than 10k voxels to get an adequate approximation.
You would need to run that numerical approach for several combinations of major radius, minor radius and mass to see how (and if) it affects the direction of the gravity. (Also, if all the tried combinations show that gravity at A always points at the same direction, it probably still isn't proof that it couldn't be reversed with some untested extreme parameters...)
Assume without loss of generality that A lies on the x-axis and B's y-coordinate is zero.
I didn't quite understand. Wouldn't that mean that both A and B are on the x-axis? Perhaps you meant that A is on the x-axis and B on the y-axis?
It's indeed hard to see when the coordinate system is not drawn. The origin is the center of the torus and A is at (d-R,0,0), while B at (d,0,R). The z-axis passes through the hole in the torus, being perpendicular to the plane of its largest circle.
Here's my numerical approximation approach. It's assuredly buggy it minor ways (don't try to stick the point you want a force for directly on the torus's surface since you end up with zero distance leading to massive absurd gravitational forces), but I think it's otherwise basically accurate.
Language: python
## Torus major radius
majorRad = 10
## Torus minor radius
minorRad = 1
## Number of rings of the torus to create
circumDivisions = 1000
## Number of points in each ring
ringDivisions = 100
## Mass of a point on the torus
# (so total mass = torusMass * circumDivisions * ringDivisions)
# Compare Earth at ~10^24kg
torusMass = 10 ** 19
## Mass of the point we calculate gravity for
targetMass = 1
## Gravitational constant
G = 6.67428 / (10 ** 11)
## Point to calculate gravity for
targetCenter = (0, 8.9999, 0)
import math
points = []
for i in xrange(circumDivisions):
theta = i / float(circumDivisions) * math.pi * 2
for j in xrange(ringDivisions):
phi = j / float(ringDivisions) * math.pi * 2
temp = majorRad + minorRad * math.cos(phi)
x = temp * math.cos(theta)
y = temp * math.sin(theta)
z = minorRad * math.sin(phi)
points.append((x, y, z))
gravyVector = [0, 0, 0]
for point in points:
delta = [point[i] - targetCenter[i] for i in xrange(3)]
distanceSquared = sum([d ** 2 for d in delta])
force = G * torusMass * targetMass / distanceSquared
distance = math.sqrt(distanceSquared)
# Normalize delta to get direction
direction = [d / distance for d in delta]
for i in xrange(3):
gravyVector[i] += direction[i] * force
for i in xrange(3):
gravyVector[i] = gravyVector[i] / len(points)
if abs(gravyVector[i]) < 1:
# Cleaner display
gravyVector[i] = 0
# Approximate gravitational force on Earth's surface is 5.5 * 10^9 newtons.
print "An object at",targetCenter,"would experience a force of",gravyVector,"newtons in X, Y, and Z"
I made the torus have approximately the same mass ratio to a human as the Earth has. Some sample output was here, but I redacted it after realizing my positions were all off ([9, 0, 1] is not on the torus). Here's some updated results:
An object at (0, 0, 0) would experience a force of [0, 0, 0] newtons in X, Y, and Z
An object at (0, 0, 0.001) would experience a force of [0, 0, -672.48108917887703] newtons in X, Y, and Z
An object at (0.001, 0, 0) would experience a force of [336.24055346705705, 0, 0] newtons in X, Y, and Z
An object at (8.9990000000000006, 0, 0) would experience a force of [6681569642.6716862, 0, 0] newtons in X, Y, and Z
An object at (11.000999999999999, 0, 0) would experience a force of [-6687714835.3757915, 0, 0] newtons in X, Y, and Z
An object at (10, 0, 1.0009999999999999) would experience a force of [-4100004.7859800849, 0, -6685114396.9591808] newtons in X, Y, and Z
Thus, the center point is stable in Z but unstable in X and Y -- no surprises there. Also, no matter where you stand on the ring, the primary force will be "down" to the ground. When standing on the "top" of the ring there's a horizontal force pulling you towards the torus's center of mass, but it's ~1500 times weaker than the downward force. Overall the downward force only varies by about a tenth of a percent for this torus.
Pyrel - an open-source rewrite of the Angband roguelike game in Python.
I had an idea: If you are calculating the gravity of a spherical object, the radius of the sphere doesn't really matter (as long as it's smaller or equal to the distance between the center of the sphere and the test point). In other words, if you eg. wanted to calculate the gravity on the surface of Earth, you would get the same result if you assumed that Earth was compressed into a point (retaining its mass) but the test point was still at the same distance from this center (in other words, the old radius).
It probably works the same with the torus, at least from the point of view of A (and perhaps even B): If we shrink the minor radius of the torus (but retaining the torus' mass) without moving A, the result would still probably be the same. We could thus shrink the minor radius to zero, in which case our problem has been reduced to a 2-dimensional one: Now we only have to calculate the gravity of a circle (at a distance of the old minor radius) rather than torus. This should remove at least one of the integrals (if not two, I'm not completely sure), making the problem significantly simpler.
Now, I don't know if the same applies to measuring the gravity at B, though.
(Edit: Of course to be completely certain of this, it would have to be proven, and this proof would be solving the volume integral and seeing that it reduces to the circle problem, which wouldn't be any easier than the original problem. I don't know if there's any other way of proving this. Of course we could just assume that the minor radius does not matter and could be zero, and that only the distance from A to the center of the torus ring matters.)
Well, for the small radius much smaller than the larger one, you could approximate it to a ring. In this case, it's clear that the gravity in A points to the surface of the torus.
Take, for example, a point inside the ring that's not the center and draw a line through it, measuring the length of the two segments d1 and d2. If you rotate this line by an angle theta, you'll get two lines inscribed in a circle with an angle theta between them. These lines enclose portions of the ring in each side, the mass of these portions is proportional to the length of the arc, which, for small enough theta, can be viewed as proportional to d1 cos(theta) for one side and d2 cos(theta) on the other. Since the gravitational force is proportional to the inverse of the square of the distance:
F1 is proportional to cos(theta)/d1 and F2 to cos(theta)/d2. Thus, if d1 <d2>F2 and the largest attraction is on the closest side, so the resultant gets to the surface. So, the idea of things getting trapped in the center of the torus makes no sense xD.
For point A this is true, for point B it is not. The distance between B and the circle on the Z axis definitely matters.
The distance from B to the circle doesn't change. Only the minor radius of the torus changes. Does that change the gravity at B?
I meant to bit about it being reduced to a 2 dimensional problem. So you're right, it doesn't change the gravity at B.
Also, p4wn3r, I don't see how the ratio between the big and large radii matters. Even if there is little (or no) difference between the big and small radius, the reduction to a ring changes the gravity for no point.
Here's my numerical approximation approach. It's assuredly buggy it minor ways (don't try to stick the point you want a force for directly on the torus's surface since you end up with zero distance leading to massive absurd gravitational forces), but I think it's otherwise basically accurate.
I think there's a mistake in the program (and what you wrote in parentheses should give a hint of why).
You are subdividing the torus into small sections and, effectively, shrinking each section into a point. However, rather than each such point being at the center of the section, where it should be, you are putting the points on the corner points of the sections. This effectively means that you have tiny black holes on the surface of the torus, which is why you can't put anything directly on the surface of the torus without getting an infinite gravity.
As said, the points should be in the centers of the volumes you are subdividing the torus into, not on their corner points.
Perhaps external references are off topic in this thread, but the gravitational field of a torus is discussed here: http://www.mathpages.com/home/kmath402/kmath402.htm
I recommend these math pages for interesting discussions on other topics in math and physics too.
I had an idea: If you are calculating the gravity of a spherical object, the radius of the sphere doesn't really matter (as long as it's smaller or equal to the distance between the center of the sphere and the test point). In other words, if you eg. wanted to calculate the gravity on the surface of Earth, you would get the same result if you assumed that Earth was compressed into a point (retaining its mass) but the test point was still at the same distance from this center (in other words, the old radius).
It probably works the same with the torus, at least from the point of view of A (and perhaps even B): If we shrink the minor radius of the torus (but retaining the torus' mass) without moving A, the result would still probably be the same. We could thus shrink the minor radius to zero, in which case our problem has been reduced to a 2-dimensional one: Now we only have to calculate the gravity of a circle (at a distance of the old minor radius) rather than torus. This should remove at least one of the integrals (if not two, I'm not completely sure), making the problem significantly simpler.
Scepheo wrote:
Also, p4wn3r, I don't see how the ratio between the big and large radii matters. Even if there is little (or no) difference between the big and small radius, the reduction to a ring changes the gravity for no point.
It does change, if it didn't, all integrals wouldn't depend on the small radius, and they do.
When you calculate gravity from the Earth, you can reduce it to a point because of several symmetric simplifications. Gauss's law (it's mostly for electromagnetism, but it applies to gravitation with some changes, since the nature of the forces is, for this case, the same) states that if we trace an enclosed surface, the gravitational flux through that surface (that is, the integral of the scalar product of the field to the differential of area of the surface) depends solely on the amount of mass inside that surface.
Because of that, if you're at distance d, trace a sphere with radius d centered at the earth's center. The gravitational flux depends only on Earth's mass, but if we assume that it's a perfect sphere, we can imply that the modulus of gravitational field is the same in all directions (and even, parallel to the differential of area of our d-radius sphere). Thus, the whole integral can be expressed as (area of sphere)*(grav field). Calculating the field this way, it'll be the same as a punctual mass.
Now, look at the torus, it doesn't have said symmetry, even from A. Trace a sphere centered at the torus's center. We can certainly see that all points in the same latitude have symmetry in relation to the field's modulus, but to simplify our calculation we need to have symmetry in the entire surface. So, this reduction doesn't work.
Ah, and another thing about the program: It only places the point masses on the surface of the torus rather than filling it. In other words, what you have is a hollow torus, rather than a filled one. (On the other hand, I don't know if this makes any difference on the results. It might not. Filling the torus would also pose problems because you would have to fill it evenly, or else you would be simulating a torus with uneven density. Again, I don't know if it would change the results.)
As discussed, it might be enough to have the minor radius as zero, and hence just create a circle of massive points. (However, rigorous mathematical proof of this would be complicated.)
p4wn3r wrote:
Now, look at the torus, it doesn't have said symmetry, even from A. Trace a sphere centered at the torus's center. We can certainly see that all points in the same latitude have symmetry in relation to the field's modulus, but to simplify our calculation we need to have symmetry in the entire surface. So, this reduction doesn't work.
So my idea of reducing the minor radius of the torus to zero to simplify calculations doesn't work after all? Bummer. It really makes things complicated.
Does it also mean that to approximate the gravity numerically, you would have to fill the entire torus evenly with point masses (iow. just filling the surface isn't enough)? Filling a torus evenly with points (so as to get even density) is not a trivial problem in itself...
I apologize for "spamming" a bit here, but I was thinking about the numerical approach program.
Derakon wrote:
Here's my numerical approximation approach.
Basically what you did was to subdivide the torus into thin pyramids, where the base of the pyramid is on the surface of the torus and the apex is at the central major-radius circle or the torus. However, the mistake is putting the mass point on the base of the pyramid (ie. on the surface of the torus) rather than at its center.
Of course putting the mass point on the center of the pyramid is probably not enough, as it probably doesn't accurately represent the gravitation of said pyramid. You would have to further subdivide the pyramid into sections from the base towards the apex and put mass points at the center of each such polyhedron. (Ok, technically speaking they are not polyhedrons because two of the faces are slightly curved, but it's close enough.) The thing is, the points cannot have equal mass, or else the density of the torus would not be even. The mass of each point would have to be proportional to the volume of the polyhedron the point mass represents. This volume can probably be approximated with some linear formula.
Overall, it's pretty complicated.
An approximation of even density may be enough for the purpose (because the whole thing is an approximation, after all). However, there should be an approximation nevertheless. If the points are not distributed evenly inside the torus (as they would not be if you do as I suggested, ie. subdivide the "pyramids" even further into polyhedrons) then their mass has to be inversely proportional to the local density to account for this, else we get a torus that is significantly denser deeper inside than closer to the surface. (As said, I don't know if this changes the results, but it might.)
Yeah, here's a first pass where the density is higher inside the torus than at its surface. I did this straightforwardly by basically generating several toroidal shells of linearly increasing minor radii.
I won't post the code since it's basically the same as before; just a different point generation system and some wrapping to let me generate results for a range of toruses. Instead of printing the gravity vector for each location, I'm just printing the nonzero components and doing some comparisons.
Some takeaways from this approximation:
* Thickness of the torus has little impact on absolute gravitational pull; the difference for the "inside" values between the thickest and thinnest toruses is only 1%
* As the torus gets thinner, the downward pull at the "inside" location becomes stronger. As expected.
* The same thing happens to a very slightly greater degree at the "outside" location.
* As the torus gets thinner (larger major:minor ratio), it becomes easier to stand on top of the torus as the sideways pull is weaker relative to the downwards pull.
Pyrel - an open-source rewrite of the Angband roguelike game in Python.
Yeah, here's a first pass where the density is higher inside the torus than at its surface. I did this straightforwardly by basically generating several toroidal shells of linearly increasing minor radii.
Are you putting the mass points in the center of the subdivided volumes or on their corners? Because, as said, if you are putting them on the corners the results will be skewed for test points very close to the surface of the torus (because they will be too close to your small "black holes" and thus experiencing invalidly strong gravitation). (The solution to this ought to be rather simple: The minor radius should be half-way between the subdivision edges.)
Anyways, the results might still be invalid nevertheless due to the distribution of mass points not being even throughout the torus. You shouldn't underestimate the difference in density that you get if you simply make concentric (minor radius varying) toruses of mass points, each one having the same amount of points, each having the same mass: The density of the inner toruses will be significantly higher than the outer ones, which might skew the results.
The solution to that should also be relatively simple: You just have to make the mass of the points relative to the minor radius, rather than all the points having the same mass. I think that the mass should be proportional to the square of the minor radius, if I understand correctly.
(Of course there's another problem: With that point distribution algorithm the points will be less densely distributed on the outer rim of the torus than on the inner rim, again causing an uneven distribution. This becomes more significant as the minor radius approaches the major one.)
Edit: I made a rendering of a cross-section of a torus showing the point distribution discussed above:
The points are significantly more densely packed in the center than closer to the surface.
Yes, I'm fully aware of how this isn't an even distribution, as well as the limitations of having a low-resolution approximation of a torus. I don't really care though; the goal was to get an approximate idea of the gravity gradients at various points and I think this simulation is entirely adequate for that purpose. If you'd like to tweak the point distribution or the mass of the points, I can send you the script I wrote.
Pyrel - an open-source rewrite of the Angband roguelike game in Python.
Yes, I'm fully aware of how this isn't an even distribution, as well as the limitations of having a low-resolution approximation of a torus. I don't really care though; the goal was to get an approximate idea of the gravity gradients at various points and I think this simulation is entirely adequate for that purpose. If you'd like to tweak the point distribution or the mass of the points, I can send you the script I wrote.
The problem is that it's not testing a torus with even density, and it has not been demonstrated that the changing the density distribution does not alter the result.
Yes, I got that the first time you said it. In fact, I knew that before I even posted my second set of results. I'm assuming that the non-constant density does not have a significant impact, without any basis for that assumption other than gut intuition. As I said, if you feel that strongly about the matter you're welcome to modify the script to use non-constant point masses or a variable distribution. Or to solve the triple integral.
Pyrel - an open-source rewrite of the Angband roguelike game in Python.
I couldn't find a software that could do a triple integral on something that's not a box, so I implemented one myself (probably not as precise as it could be): http://pastebin.com/k8RTmBsR
This code is able to integrate n-dimensional integrals, experienced programmers should be able to quickly modify it to calculate the gravity field at any point. The present code outputs:
Point A
Ratio 0.05 : ( 0.2795 , 0.0000 , -0.0000 )
Ratio 0.10 : ( 0.5272 , -0.0000 , -0.0000 )
Ratio 0.15 : ( 0.7439 , 0.0000 , -0.0000 )
Ratio 0.20 : ( 0.9320 , -0.0000 , 0.0000 )
Ratio 0.25 : ( 1.0927 , -0.0000 , 0.0000 )
Ratio 0.30 : ( 1.2260 , -0.0000 , -0.0000 )
Ratio 0.35 : ( 1.3317 , -0.0000 , -0.0000 )
Ratio 0.40 : ( 1.4095 , 0.0000 , 0.0000 )
Ratio 0.45 : ( 1.4585 , 0.0000 , 0.0000 )
Ratio 0.50 : ( 1.4778 , 0.0000 , -0.0000 )
Ratio 0.55 : ( 1.4661 , 0.0000 , 0.0000 )
Ratio 0.60 : ( 1.4228 , -0.0000 , -0.0000 )
Ratio 0.65 : ( 1.3465 , -0.0000 , -0.0000 )
Ratio 0.70 : ( 1.2377 , -0.0000 , -0.0000 )
Ratio 0.75 : ( 1.0949 , 0.0000 , 0.0000 )
Ratio 0.80 : ( 0.9189 , 0.0000 , 0.0000 )
Ratio 0.85 : ( 0.7124 , 0.0000 , -0.0000 )
Ratio 0.90 : ( 0.4792 , 0.0000 , -0.0000 )
Ratio 0.95 : ( 0.2233 , -0.0000 , 0.0000 )
Point B
Ratio 0.05 : ( -0.0300 , -0.0000 , -0.3079 )
Ratio 0.10 : ( -0.0981 , 0.0000 , -0.6236 )
Ratio 0.15 : ( -0.1916 , 0.0000 , -0.9412 )
Ratio 0.20 : ( -0.3034 , -0.0000 , -1.2608 )
Ratio 0.25 : ( -0.4283 , -0.0000 , -1.5823 )
Ratio 0.30 : ( -0.5622 , 0.0000 , -1.9055 )
Ratio 0.35 : ( -0.7019 , 0.0000 , -2.2300 )
Ratio 0.40 : ( -0.8442 , 0.0000 , -2.5553 )
Ratio 0.45 : ( -0.9870 , -0.0000 , -2.8808 )
Ratio 0.50 : ( -1.1280 , -0.0000 , -3.2058 )
Ratio 0.55 : ( -1.2656 , -0.0000 , -3.5296 )
Ratio 0.60 : ( -1.3981 , -0.0000 , -3.8511 )
Ratio 0.65 : ( -1.5244 , -0.0000 , -4.1698 )
Ratio 0.70 : ( -1.6433 , 0.0000 , -4.4846 )
Ratio 0.75 : ( -1.7543 , 0.0000 , -4.7950 )
Ratio 0.80 : ( -1.8568 , -0.0000 , -5.1000 )
Ratio 0.85 : ( -1.9502 , 0.0000 , -5.3991 )
Ratio 0.90 : ( -2.0346 , -0.0000 , -5.6915 )
Ratio 0.95 : ( -2.1094 , -0.0000 , -5.9767 )
C++ has an awful habit of printing -0.0000 . I told it to evaluate the zero values to check its correctness. The components of the vectors, when multiplied to G*\mu*R, where G is the gravitational constant, \mu the torus density and R the largest radius, will yield the acceleration caused by gravity at that point. We can see that the torus world is very unstable at B, with some nasty pulls towards the center. I plotted A's x components and could get a great fit with a parabola:
So on a toroidal world gravity would still always point "down" at point A regardless of the proportions of the torus, which is an interesting result. It could, in principle, make a toroidal world plausible.
Of course gravity would be "normal" only at the points which are closest or farthest from the center of the torus. Gravity gets slanted when you are anywhere else, seemingly quite significantly at points. This would mean that on large parts of the map gravity would be heavily oblique with respect to the ground, causing interesting results (and possibly making life very difficult if not impossible).
I'm assuming that if the surface of the planet was completely smooth (iow. it's a mathematical torus) and we put a ball on point B, it would roll towards A due to gravity (and finally settle on A if friction stops it eventually). Would this be a correct assessment?
I cannot guarantee that the integrals have finite form primitives, perhaps someone could fill some values and evaluate them numerically to see the direction the vectors point?
EDIT: Does anyone want to know the math behind this? I think it's very annoying...
EDIT 2: I keep confusing R with r!!!!
In the last two integrals, r^2 (sin fi - 1) read r^2 sin fi - R^2. And in the third, r^3 (sin fi - 1) = r^2 ( r sin fi - R)
