Co- authors Wanted for Journal Paper (related to "Jello-O... " thread)

Good idea. If you are sure the same force field is produced then I doubt any shape, not just the eggoid we have assumed, will be lower potential than the sphere. Perhaps the sphere is stable under push gravity that acts thru out the volume and we can abandon this effort. MacM will say "I told you so" but now at least his version has passed one hurtle or two thanks to our efforts.

 

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Here is a table of results from my method. For all of these I set y==0 since, by symmetry, the y-directed forces sum to zero and all planes containing the z axis are equivalent. (I did, of course, integrate over y and verify that it summed to zero, but I just don't want to clutter up the table). I randomly generated 0 < D < 1 and randomly picked a point inside the eggoid in the s,z plane. I then calculated a^2, the UniKEF force and the classical force. Note that the UniKEF force agrees with the Newtonian force in all cases tested.

Aer, could you pick a few of these and do your integration and see if you get the same results?

-Thanks
Dale

<font face=courier new>..D........(s,z)........a^2...........UniKEF.............Newton

0.662 ( 0.302, 0.309) 0.968756 (-1.32108,-0.47045) (-1.32108,-0.47033)
0.077 (-0.101, 0.064) 0.999556 ( 0.42357,-0.17149) ( 0.42355,-0.17148)
0.619 ( 0.274,-0.519) 0.972514 (-1.04041, 3.10793) (-1.04041, 3.10798)
0.386 ( 0.480,-0.555) 0.989020 (-1.89087, 2.89216) (-1.89086, 2.89215)
0.471 ( 0.701, 0.354) 0.983790 (-3.05278,-0.96194) (-3.05279,-0.96194)
0.887 (-0.018, 1.394) 0.946003 ( 0.09156,-3.44037) ( 0.09155,-3.44037)
0.361 ( 0.496, 0.502) 0.990376 (-2.16689,-1.60995) (-2.16694,-1.60994)
0.857 ( 0.802,-0.145) 0.949311 (-3.24744, 1.34120) (-3.24742, 1.34122)
0.618 ( 0.286, 1.023) 0.972599 (-1.35450,-2.97874) (-1.35449,-2.97872)
0.462 ( 0.748, 0.257) 0.984389 (-3.22297,-0.60979) (-3.22297,-0.60978)
0.335 (-0.170,-0.364) 0.991694 ( 0.68803, 1.98946) ( 0.68803, 1.98946)
0.042 (-0.037, 0.735) 0.999868 ( 0.15620,-3.00187) ( 0.15618,-3.00184)
0.350 ( 0.102, 0.270) 0.990945 (-0.43665,-0.67434) (-0.43664,-0.67435)
0.496 ( 0.200, 1.058) 0.982074 (-0.93214,-3.31824) (-0.93215,-3.31828)
0.690 ( 0.166, 0.058) 0.966201 (-0.69830, 0.58011) (-0.69833, 0.58012)
0.609 ( 0.483, 0.799) 0.973358 (-2.23531,-2.31423) (-2.23532,-2.31423)
0.275 ( 0.737, 0.038) 0.994379 (-3.09454, 0.10446) (-3.09454, 0.10446)
0.569 ( 0.712, 0.559) 0.976618 (-3.19932,-1.61419) (-3.19929,-1.61413)
0.850 ( 0.826, 0.026) 0.950072 (-3.47734, 0.59403) (-3.47733, 0.59402)
0.356 (-0.785,-0.458) 0.990636 ( 3.14358, 2.32150) ( 3.14357, 2.32148)
0.570 (-0.212,-0.334) 0.976539 ( 0.83911, 2.16302) ( 0.83912, 2.16302)
0.533 ( 0.092,-0.669) 0.979390 (-0.34538, 3.74614) (-0.34538, 3.74594)
0.540 (-0.781, 0.707) 0.978863 ( 3.55477,-2.19351) ( 3.55469,-2.19348)
0.156 ( 0.233, 0.658) 0.998180 (-1.00070,-2.49539) (-1.00072,-2.49539)
0.355 (-0.896, 0.034) 0.990688 ( 3.76523, 0.14697) ( 3.76521, 0.14679)
</font>

 

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I could and do. However, that is to acknowledge that to do so is speaking from logic and the gut and not mathematics. What has been done here is important confirmation.

 

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Well it looks like all of your forces point towards the centroid. Mine do not, which is why I have a suspicion there is some error, maybe just a typo. Can you expand on how you did your integration? For instance, I do it the following way (using basic summation instead of integration):

Fs = &int; &int; (L1-L2) cos &beta; cos&alpha; d&beta; d&alpha;

Fz = &int; &int; (L1-L2) sin &beta; cos&alpha; d&beta; d&alpha;

L1 and L2 being functions of &alpha; and &beta;.

 

<font face=times new roman>I did it slightly different than you did. I used an algebraic approach while you used a geometric approach. First, I started with the equation for the eggoid:

(1) S²/A² + Z²/(1 + D Z) == R²

I solved (1) for S as a function of z and then integrated

(2) V == ∫ π S[z]² dz

to get an expression for the volume. I then set V = 4/3 π R³ and solved (2) for A² to get an expression for the A² needed to get an eggoid with the volume equal to a sphere of radius R.

I then calculated the z position of the centroid as follows (s position == 0 by symmetry):

(3) C == (∫ π z S[z]² dz)/(∫ π S[z]² dz)

At this point I switched from cylindrical coordinates to Cartesian coordinates. Specifically, in (1) I made the substitution S² = X² + Y² giving.

(4) (X² + Y²)/A² + Z²/(1 + D Z) == R²

I then wrote the parametric equation for a line in Cartesian coordinates where bold indicates a vector quantity:

(5) x(t) == x0 + t v
where x0 == (s0, 0, z0)
and v == (cos[φ] sin[θ], sin[φ] sin[θ], cos[θ])

Note that by symmetry there is no loss in generality using this definition for x0. Also note that v is a unit vector and that as θ varies over [0,π/2] and φ varies over (-π,π] we obtain the set of all lines passing through x0. Finally, note that t is the directed distance between x(t) and x0.

For any x0 on the interior of the eggoid and for any v there are two real solutions, t1 and t2, for the simultaneous equations (4) and (5). Since t is a directed distance one solution will be positive and the other will be negative. Therefore the expression

(6) f[s0,z0,θ,φ] == (t1+t2)v

is the UniKEF force component in the v direction at x0. The total UniKEF force at x0 is given by the following integration over all v:

(7) ∫ ∫ f[s0,z0,θ,φ] sin[θ] dφ dθ

Unfortunately, (7) could not be evaluated analytically, so I had to evaluate it numerically for specific values of s0, z0, and D (I always set R=1 and calculated A from the solution for (2) described above). As anticipated by symmetry, when I evaluated (7) numerically I found that the y component of f was always zero, so I only reported the x and z components (or s and z depending on how you want to look at it).

-Hope this helps
Dale

PS Sorry about having to use [ ] for function arguments instead of ( ). For some reason smileys are processed first so ( ) with θ or φ is interpreted poorly, e.g. sin(φ).
</font face>

 

I plotted the field map with my red force lines all made much longer and I zoomed in to the area surrounding the centroid. This is what I got.

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My lines don't go through the centroid either. I suppose that it could be due to numerical errors in the integration, but then I would not expect the left-right symmetry. I think you are probably ok.

-Dale

 

OK - I agree with everything thus far. Up to this point, we have done everything exactly the same.

OK - here I don't think I agree. Yes it is true that θ varies over 0,π and φ varies over -π,π when you are talking about locations in spherical coordinates. However, when dealing with the flux integration from all directions, the range for both should just be 0,π.

Here is the reason why: First let's imagine why the first set of intervals are used to describe a sphere in spherical coordinates. If you draw a curved line for a certain radius, r over the range of θ, 0,π - you will have a vertical half circle. Next if you rotate this half circle about the center over the range of φ, -π,π - you're left with a hollow sphere. Integrating over &rho;, the radius produces a solid sphere.

Now when dealing with the flux from all directions, when you solve for t1, that should be in the direction &theta;,&phi;. When you solve for t2, that should be in the direction &theta;+&pi;,&phi;+&pi;. So when you get to &theta;=&pi; and &phi;=&pi; you will have solved for the entire 2&pi; range. Do you follow what I am getting at?

 

Oops, you are absolutely correct. Since each line consists of a positive ray and a negative ray I integrate over half the sphere, not over the whole sphere. My code is correct (integrating over half the sphere), my post was in error. I will go back and fix my post as soon as I finish here. The only difference is that I integrated over a different half sphere than you did. My integration is φ varies over -π,π and θ varies over 0,π/2.

-Dale