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

Discussion in 'Physics & Math' started by Billy T, Sep 16, 2005.

1. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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It would seem that way*, but I do not completely undestand your drawings yet. I have a lot of problems with the appearance that your new eggoids relative to the sphere as it differs from your

and Dale's earlier eggoids in that part of the new eggoid appears to be expanded for nothing if the expansion is radial and part of the sphere has completely disappeared. some of this is due to the fact that the eggoid is not with centroid at the origin, but not all. please do not misunderstand. I like its shape better, but I am not sure it has the same volume as the sphere and not sure how to write the dw expression for values of my "z pancakes" especially for z locations that do not have any sphere to expand from.

Have you followed my "z pancakes" & "z annulus" ideas? And my idea of three integrations, I1, I2 & I3? I1 was the integration for fixed flux and z along the radial expansion (or contraction) for dw. I2 was the results of I1 along all values of z. I3, if needed, was for all flux angles, now called &beta;
_____________________________

* I have long agreed and argued that the force is purely centrally directed in the case of the sphere, and thus basically so in the eggoid case, but if the eggoid is contracted everywhere from z = zc (the crossing point's z) towards one end and expanded from the sphere from zc to the other end, then one region is doing work and the other is having work done on it, so it is not too important if the force is everywhere nearly towards the centroid. Water goes to a lower energy state despite the force on it being always towards the Earth's center.

Last edited by a moderator: Nov 24, 2005

3. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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These are two points. thus they define a line, not a plain or a unique cross section. I am still confussed.

I pivot (or rotate) about lines, not points. I am not with you at all. Can you try again? thanks.

5. ### DaleSpamTANSTAAFLRegistered Senior Member

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For the sphere we have already shown that it is the same form as classical gravity, which is conservative. As long as the curl of the any vector field is zero then the field is conservative. I cannot see how we can possibly get a non-zero curl with UniKEF. Push or pull, it is irrotational.

When we evaluate the vector field over the eggoid we can also take a first-order numerical approximation to the curl. If this is reasonably small then we can safely assume it is conservative.

You are right. I said it backwards. The gradient of the scalar potential gives the vector force.

Billy, the problem is that I do not understand virtual work at all. I have never used it personally, it is not in any of my textbooks, and the few online references I was able to dig up only applied to solid objects (with a stress/strain relationship). Your obsession with virtual work is not helpful at all; from everything I can see it is not relevant and even if it were I don't understand it enough to use/adapt it. If you want to modify the virtual work method to the present situation then simply tell me explicitly and exactly the integral you want me to evaluate and I will gladly do so, but I am not going to be able to derive a virtual work integral myself.

I don't think that showing this is as difficult as you think it is, and if we don't show it then we will have a much worse problem anyway. You mentioned that you want to look at extreme deformations. For a small D we can probably reasonably approximate the deformation paths as purely radial, but with large D that approximation will no longer hold. That leaves us with an infinite number of deformation paths to consider. On the other hand, if the field is conservative then we only have to consider the beginning and the end states. All possible deformation paths will have the same work. I think it is easier to check for conservatism than to evaluate an infinite number of path integrals. Besides, if we find that UniKEF is non-conservative then there is no need to worry about path integrals since gravity is demonstrably conservative.

-Dale

7. ### DaleSpamTANSTAAFLRegistered Senior Member

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Yes, there is a division by D. This is not an error in your code, this is correct and is consistent for all 3 eggoid equations. Because of that I am a little suspicious about the force results for small D, but since we know the analytical form of the sphere we should be able to detect any gross errors.

To get an accurate formula for small D I think we will have to do a Taylor series expansion in D.

Sorry, I am getting a little behind in this thread. Hopefully I will have a little more time tomorrow evening.

-Dale

8. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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yes but the external potential of a gravaty sphere is for a test mass at the location of the potential. This is not our problem. We want the net energy reqired to deform the sphere A SINGLE OBJECT - NO TEST MASS. If we must put energy in to defrom it it will not deform by the path we assumed.
With numerical results that is technically true, but if for various Ds there is a smooth curve on a "work done" vs D graph and it is always positive (or negative) for all Ds considered, I think most people will agree that we need not consider all possible Ds.
yes but you are missing the point of my Roman aqueduct example. All possible means for the water in the aqueduct to get to the valley floor 50 meters below will give the same change of potential for a Kg of water 50 meters lower, but it will not go there as it must first go "up hill" to get over the aqueduct wall. Thus it does not matter that the valley floor is a lower potential energy state than the village the water actually goes to.

I am not every asking for a comparison of two significantly different states. Initally I want D to be as near to zero as the numerical accuracy permits. Then I will want a little larger D2 etc. but agree that we need to consider only a small set of Ds and show, perhaps graphically, that the relationship between two nearly equal but significantly non zero Ds, is one of continued down hill nature if we find that the first D considered indicates that some small deform of the sphere will occur.

Yes, I only ask for help with the math. (I will again get an expression for dw, like the one you accepted before.) At present I need time to think how to handle the new eggoid. It differes from the one that was either expanding of contracting in that parts of it seem to come from "nothing" See my earlier comments relative to calculating the work prior to shiftng the centroid even if wrong just to learn how to do the work calculation. Again I don't care what we call it but we must evaluate the work done as the defomation is done. I am not sure what use a "potential of a test mass" has to do with a deforming single body problem. This is not a two body broblem. We are not bringing in a test mass from infinitely far away. This is how the normal (negative) gravitational potential is defined. In all cases, however, the change in potential is computed by the work done during the change. We need not call it virtual work or Billy's T method etc. just need to see if we can compute it for a sphere deforming by a continuous path to an assumed new shape, makying sure not point on the path is "up hill" (requires external source of energy)

-Dale[/QUOTE]

Last edited by a moderator: Nov 24, 2005
9. ### AerRegistered Senior Member

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2,250
Thanks.

I didn't think it was an error, per say - just that it was in there somewhere. As it turns out the it was in the A equation. Luckily we know that A should be equal to 1 when D is 0, so I just made an if/else statement so that it will be able to handle D=0. Everything works fine and I get basically the same result as I did for D=.001. I don't think there is any reason to worry about setting D = 0.

10. ### AerRegistered Senior Member

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It's just another way of saying they will not cancel out (i.e. they add upon each other) or:
Same thing. I suppose compound can have the wrong implied meaning as you may tend to think of powers with compounding, but I just meant for the meaning of the word in the most generic sense: add together.

You are forgetting though that the flux lines have to be considered from all directions. As stands you are only assuming an eggoid disk. This is just the same as only assuming a circle and saying the result applies for a sphere. Is that what you want? You seem to get all over MacM's case about this very same thing that you are now proposing we do.

Yes, but to compute this force, you still must integrate over &alpha;, not just &beta;. I am sorry but I don't understand what is so hard to comprehend here.

You have a point in the eggoid. You draw a line from the centroid to the said point. Now if you draw a plane parallel to the z axis in which the line lies within this plane, then you have the cross-section at &alpha;=0. With this cross section, you integrate over &beta; which I assume you understand very well.

Now before I said you rotate this cross-section about the point assuming you knew that you rotate about a line through the point that is parallel to the z axis. anyway, you rotate it in such a way and the angle of rotation is &alpha;. Notice that you will have a new cross-section through the 3D eggoid. This cross section will have a component force in z and s as well as a component force perpendicular to the zs plane. This component, however, can be ignored because it will cancel out. However, you still must account for the z and s components by integrating over &alpha;. It is not as simple as just multiplying by a constant as you have stated before. I just finished incorporating this integration into the code.

&alpha; has a range of just &pi; just like &beta;. We are not talking about spherical coordinates here. Just basic rotations. What is so hard to understand? If we don't include &alpha; then we will get a wrong result. Most notably, the force will not fall off with r&sup2; if we discount &alpha;.

11. ### AerRegistered Senior Member

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I don't really have any idea what you are alluding to here. You speak of expansion when the volume of the eggoid is equal to that of the sphere (I calculate it in my code just to make sure, so you can sleep easy about the volumes being the same). There isn't really any expansion going on here. You also talk of the sphere disappearing... What does this mean? I suppose you could see it as part of the sphere disappearing, but then you would have to look at the other portion as the sphere reappearing... eh?

As I said, the volume is in fact the same. That was the entire point of including the A factor as to set the volumes equal to each other.

Never heard of them.

No.

so I1 is in the s direction, I2 is in the z direction, and I3 is any direction inbetween? If so, &beta; accounts for all of your I's. &alpha; accounts for directions with a perpendicular component, but note that the there are still z and s components along, because when you rotate &alpha;, you have to do the &beta; integration all over again.

12. ### AerRegistered Senior Member

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I figured it was understood by everyone that the cross-sections, all the cross-sections were parallel to the z plane. Therefore if you define a line, then you can make a unique cross-section ( unless the line is parallel to the z plane - luckily (or purposefully) all of the lines I've defined have not been parallel to the z axis as they are by definition parallel to the s axis (a rotation of &alpha; is in the s plane.. as I mentioned before) ).

13. ### AerRegistered Senior Member

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2,250
Here is an image that maps out the force gradient for a path from the surface to centroid:

These forces were computed with a full integration over &alpha; and &beta;

N corresponds to the number of &beta; increments and M to the number of &alpha; increments. N & M were set to 40 which amounts to 800 (ha, 1600) numerical convergences per point and there are 14 points in the image. This only took about 7 seconds, but that will become a lot longer if we have to integrate over all points in the sphere. Hopefully that is not what we plan on doing?

Last edited: Nov 24, 2005
14. ### DaleSpamTANSTAAFLRegistered Senior Member

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If the UniKEF force is non-conservative then there is no such thing as a work vs D graph. In fact, it doesn't even make sense to talk about a work vs D for a non-conservative force. The reason is that there are an infinite number of possible deformation paths that will take you from a sphere to any given D. If the force is non-conservative then the work will depend on the details of the path, not just the endpoint, so there will be an infinite number of work values for a given D value. Therefore, for a non-conservative force, work is not a function of D!

-Dale

15. ### DaleSpamTANSTAAFLRegistered Senior Member

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Excellent. I was hoping that it would be well-behaved for small but non-zero D. It looks like it is, at least for D on the order of .001. If we need smaller D and find that it becomes numerically unstable I will try to do a Taylor series expansion around D=0, but hopefully it won't be needed.

-Dale

16. ### DaleSpamTANSTAAFLRegistered Senior Member

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Very nice result! That is interesting how the force does not always point towards the centroid. By the way, does your path from the surface to the centroid curve, or is that just a visual illusion caused by the arrows?
I think what Billy is planning on is actually worse. I think he wants to calculate the work as every point in a sphere deforms along all possible paths to every point in an eggoid.

I should be caught up today. If our results match, then I think we can assume they are correct.

-Dale

17. ### AerRegistered Senior Member

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2,250
Yes the path curves. As a side note, that circle is just where I suspect the centroid is. Unfortunately the centroid equation is going to be different for the 3D eggoid than the 2D eggoid that you calculated.
Do you know how to compute the 3D centroid location exactly? I suspect that the closer you are to the centroid, then the force vector will point directly to the 3D centroid location. It was in that manner that I found the centroid location by trial and error plotting points such as x0=[.01,3.75] until the force vector was almost parallel to the s axis.

Perhaps he can settle for just a few select points and interpolate from there.

I am still unsure about the force calculation as I have it. However, it gives the expected result for a sphere, D=0 - so there is a chance that it is giving the correct result for the eggoid, D &ne; 0.

18. ### DaleSpamTANSTAAFLRegistered Senior Member

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The equation I gave you was the centroid for the 3D eggoid.

-Dale

19. ### DaleSpamTANSTAAFLRegistered Senior Member

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Here is my first field plot.

The eggoid is D=0.5 R=1. Unfortunately, it takes the algebraic approach I used in Mathematica much longer to evaluate than the geometric approach you used in Matlab. The 76 field points took a total of just under 35 min or approximately 27.5 s each.

-Dale

20. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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Mainly To Aer:
Have not read all carefully, but Aer is correct and finally got thru my thick skull.
I do need his &alpha; integration. The way I would put it is that at the point of interest, Xo, there is a cone of flux, not just the flux at &beta; That is I need &alpha; and &beta; to describe the rays incident on that point. I was trying to avoid a required integration by noting that an annulus of Xo points would have simple net force in any direction (and especially in the direction we planned to let it move) that is proportional to s. I.e. if the s1 annulus is twice the s2 annulus in circumference, then the force on it in any direct is twice as large. That is true but I was forgetting all of the ray incident upon the point Xo from out side of plain containing the z-axis and point Xo.

That is what you have been trying to get thru my thick skull, is it not Aer?
(We always have a little trouble communicating and I am not sure I yet follow your text, but at least the main point is now understood - I think). Thank you. This also illustrates how good it is to have three of us. When we can all agree, we probably have it correct.

21. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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I have been thinking about the use of "potential" instead of "virtual work" as the basic foundation and have also come around to Dale's (and probably Aer's) POV, I miss read Dale. He correctly stated there is no potential vs D graph. Soon I will remove remainder of this pargraph which was: although I do not agree with Dale's post that claims (as I understood it in quick skim) in a non conservative field, computing the work done by distorting over an assumed path to an assumed final shape is not possible or not a test of stability, but will not agrue this as

I now think that only two things are required to test for stability:
(1)The "full potential" (more later on what the "full" means) of the final state/configuration must be lower (for the final state to be spontaneously achived, if it is the only alternative) than the initial state (Sphere in first phase of this effort. "Second phase" starts with the slightly deformed sphere = eggoid with very small D, and test to see if the deformation continues for larger D etc. but obviously we can not do all pairs of slightly different Ds. See end of point two for what to do instead.).
And
(2)There must be some path / deformation between the initial and final states that is everywhere "down hill" (Recall my Roman aqueduct example - the water does not go 50 meters down to the valley floor, but to the village only 20 meters lower as can not climb over the "up hill" wall of the aqueduct.) This is why we need to use a small D, then one a little larger etc and pass a curve of potential vs D thru the computed points and note that it is smooth and in no danger of turning "up hill" anywhere.

Also note that I said nothing about shifting the centroid to the origin. - I think it was not necessary to be concerned with where the centroid is. Sorry if I was cause of extra work. Why I was correct to insist that both must have the same volume is now more obvious.

Thus I now agree that we need to first:
find an expression for the force throughout the volume and
then
to integrate it over the volume. If (and when?) we consider non uniform density distributions, we need to weight the force at Xo by the density at Xo when doing this integration over volume.

It is this density weighted intergal's result that I call the "full potential."

I have etched in my brain so deeply that potential is the work done on a test charge (mass in gravity case) as it is moved from the location of zero potential (usually infinitely distant) that it has been a strugle for me to agree that if we know the force every where we integrate it and throw in a spacially uniform constant that sets the zero potential at any point we like as this constant will disappear when we take the gradient to get the force back from the potential.

I think my devotion to "test charges" may have been useful as it makes the need for weighting by density in the force integration obvious (when it is not uniform). For example: Assume some mass distribution. The consider a small differential volume of it , dV, and extract this dV transporting it to infinity. There will be work done to take it away in direct proportion to the density at dV. Redefine the zero of potential to be the state with dV at infinity. Now lower dV back to its original location inside the mass distribution. The potential of the original system, wrt 0 at dV very distant, is negative and equal to the work done on the restraints of dV as dV is slowly lowered (no KE given to dV) back into its place inside the mass distribution, which is of course proportional to the density of dV.

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22. ### Billy TUse Sugar Cane Alcohol car FuelValued Senior Member

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As a "shake down test" of the potential method, if not a lot of work, Why not consider two equal-volume spherically-symmetric configurations: One a sphere or radius R1 and the other radius R2 > R1 with a hollow spherical core of radius r? Everyone will agree that the hollow sphere will distort to the solid one. It would be nice (again if not too much work) to calculate the force field in each, integrate these force fields and find / show that the hollow sphere was at a higher potential.

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23. ### DaleSpamTANSTAAFLRegistered Senior Member

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I found a simple way to speed up my computations. They now run at about 7.5 s each or under 10 min for the same 76 points.

By the way. I decided to also do the classical (Newtonian gravity) integration for a few points. It turns out that so far the UniKEF and classical forces match. I don't think there is any point in doing the rest of the computation. We already know the minimum energy state for classical gravity is a sphere, and from what I can tell UniKEF matches classical.

To make sure that I am not biasing things by picking special points I will generate a list of random points in random eggoids and calculate the classical and UniKEF force. Hopefully Aer can duplicate my results using his method.

-Dale