Possible Charged Particle Field
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Re: Possible Charged Particle Field
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Jared, I don't know about you, but I'm not script smart. I'd be more than happy to work with you to learn this app to the extent that you might need it. Helping you would help me learn it. I can't say I can teach it since there's a great deal about it I haven't begun to figure out and it's not finished yet. As Nevyn says, programming is a tool. It's more important to understand how to work with it.
I’ve got a tack for you Nevyn, but fair’s fair, you deserve credit. By expansion theory, we know charge particles are expanding. We also know, given an interval of time, a charged particles’ emissions will expand outward from the particle, as on an expanding spherical layer. Have you told us something we don’t know? I don’t think so. On the other hand, I believe this development of yours improves our understanding of relativity. Now we can say – of course! We can clearly see the both the expanding charge particle and its expanding emission field, easily adjusting for micro or macro.
I'm surprised you feel implementing the R function is a problem. I believe we already know the radius function R. All objects in the universe are expanding – accelerating  at the same rate. For Earth, the radius doubles every 19 minutes or so. Why there’s a post or two on the subject here at this site. As requested, just poking.
All radii and distances are time dependent variables – the relativistic perspective. We know that the particle radius should be time dependent, equal to a distance, r0 at some position p0 at time t0, r(t)= p(0) + r(0)*expansion(t). Normally one can overlook such details, but if you’re trying to model the universe, relativity must be included. Previously I could always claim I didn't understand relativity, but that excuse is worn pretty thin now, although you or Miles have always surprised me with things I hadn’t considered.
So the CPIM app already includes spin and a latitude function – great, it may go all the way. With all due respect, when I said you should use spin to tie all your formulas together, app aside, the physics comes first. If you’ve already built CPIM with these ideas in mind, not just a charge field but a relativistic charge field I’d say you’re doing a fine job.
Otherwise, as we're all aware, it's not quite done. It needs gravity  or this great new expansion alternative you're putting together.
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Jared, I don't know about you, but I'm not script smart. I'd be more than happy to work with you to learn this app to the extent that you might need it. Helping you would help me learn it. I can't say I can teach it since there's a great deal about it I haven't begun to figure out and it's not finished yet. As Nevyn says, programming is a tool. It's more important to understand how to work with it.
I’ve got a tack for you Nevyn, but fair’s fair, you deserve credit. By expansion theory, we know charge particles are expanding. We also know, given an interval of time, a charged particles’ emissions will expand outward from the particle, as on an expanding spherical layer. Have you told us something we don’t know? I don’t think so. On the other hand, I believe this development of yours improves our understanding of relativity. Now we can say – of course! We can clearly see the both the expanding charge particle and its expanding emission field, easily adjusting for micro or macro.
I'm surprised you feel implementing the R function is a problem. I believe we already know the radius function R. All objects in the universe are expanding – accelerating  at the same rate. For Earth, the radius doubles every 19 minutes or so. Why there’s a post or two on the subject here at this site. As requested, just poking.
All radii and distances are time dependent variables – the relativistic perspective. We know that the particle radius should be time dependent, equal to a distance, r0 at some position p0 at time t0, r(t)= p(0) + r(0)*expansion(t). Normally one can overlook such details, but if you’re trying to model the universe, relativity must be included. Previously I could always claim I didn't understand relativity, but that excuse is worn pretty thin now, although you or Miles have always surprised me with things I hadn’t considered.
So the CPIM app already includes spin and a latitude function – great, it may go all the way. With all due respect, when I said you should use spin to tie all your formulas together, app aside, the physics comes first. If you’ve already built CPIM with these ideas in mind, not just a charge field but a relativistic charge field I’d say you’re doing a fine job.
Otherwise, as we're all aware, it's not quite done. It needs gravity  or this great new expansion alternative you're putting together.
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LongtimeAirman Admin
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Re: Possible Charged Particle Field
No, I haven't shown anything we didn't already know, but I have used it in a more logical, consistent, and precise manner.
I don't know where that doubling every 19 minutes comes from, so I don't know if I can trust it. Doubling is also a relative term, so it doesn't really tell us anything about the actual radius. You also have to know what that 19 minutes applies to. You see, from a nonexpanding perspective, both distance and time expand, since time is based on distance. If the meter expands, then so does the second. This keeps the ratio the same in spite of the expansion, from an expanding perspective. I am trying to find math that fills the gap between what we see from an expanding perspective and what really happens from a nonexpanding perspective. I am not exactly in either camp, but somewhere between them. I need to step through it with great care to keep everything in its proper place.
Finding this change in radius will actually allow me to implement gravity at the same time. Since that change in radius is gravity. I just need to account for time discrepancies. I'm looking over Miles Third Wave papers for equations I can use and have found a few, but I'm still trying to figure out exactly how he calculated an initial radius for the proton. Miles states that you can't just use the equations of motions to find an older radius, or you can but you can only use it as a proportionality. I found another equation this morning, but haven't had any time to play with it.
I don't know where that doubling every 19 minutes comes from, so I don't know if I can trust it. Doubling is also a relative term, so it doesn't really tell us anything about the actual radius. You also have to know what that 19 minutes applies to. You see, from a nonexpanding perspective, both distance and time expand, since time is based on distance. If the meter expands, then so does the second. This keeps the ratio the same in spite of the expansion, from an expanding perspective. I am trying to find math that fills the gap between what we see from an expanding perspective and what really happens from a nonexpanding perspective. I am not exactly in either camp, but somewhere between them. I need to step through it with great care to keep everything in its proper place.
Finding this change in radius will actually allow me to implement gravity at the same time. Since that change in radius is gravity. I just need to account for time discrepancies. I'm looking over Miles Third Wave papers for equations I can use and have found a few, but I'm still trying to figure out exactly how he calculated an initial radius for the proton. Miles states that you can't just use the equations of motions to find an older radius, or you can but you can only use it as a proportionality. I found another equation this morning, but haven't had any time to play with it.
Nevyn Admin
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Re: Possible Charged Particle Field
I found the pin!
While the above analysis is still applicable, I have started from the wrong position. It is not volume that I should be analyzing, but surface area. My method is still the same, but we start from SA = 4Pi * r^2.
Which gives us a 1/r^2 relationship.
Therefore, to find the charge at a given distance, we use: D(d) = C(1)/d^2
The relativistic version is: D(d) = C( R( d/c ) )/d^2
Miles then went on to include gravity, which he said brings in another 1/r^2 relationship, so we end up with 1/r^4. However, I stand by my analysis and in the case of charge density, gravity must be included in the initial determination of charge density at the boundary of the emitter.
While the above analysis is still applicable, I have started from the wrong position. It is not volume that I should be analyzing, but surface area. My method is still the same, but we start from SA = 4Pi * r^2.
Radius  Area  A(r) / A(1)  A(1) / A(r) 
1  12.5663706144  1  1 
2  50.2654824574  4  0.25 
3  113.0973355292  9  0.11111111 
4  201.0619298297  16  0.0625 
5  314.159265359  25  0.04 
6  452.3893421169  36  0.0277777778 
7  615.7521601036  49  0.0204081633 
8  804.247719319  64  0.015625 
9  1017.8760197631  81  0.012345679 
10  1256.6370614359  100  0.01 
Which gives us a 1/r^2 relationship.
Therefore, to find the charge at a given distance, we use: D(d) = C(1)/d^2
The relativistic version is: D(d) = C( R( d/c ) )/d^2
Miles then went on to include gravity, which he said brings in another 1/r^2 relationship, so we end up with 1/r^4. However, I stand by my analysis and in the case of charge density, gravity must be included in the initial determination of charge density at the boundary of the emitter.
Nevyn Admin
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Re: Possible Charged Particle Field
To be clear, charge density is a function of the volume of the emitter at the time of emission. It is the emission of a volume, over a surface. However, the dropoff of that emission is only dependent on surface area. We are taking the emission of 1 dt, and watching it as it moves out into space. We don't care about the space between that emission and its source, which using volume would include, so we use the surface area equation.
Nevyn Admin
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Re: Possible Charged Particle Field
That makes perfect sense to me, Nevyn. The math even makes sense, which is a good thing since I'm not great with these kinds of maths and would have never seen these relationships the way you do.
I apologize to just be a sort of cheerleader here, but it's still interesting and I'm trying to follow along as best I can. Wish I could actually help on this one though.
I apologize to just be a sort of cheerleader here, but it's still interesting and I'm trying to follow along as best I can. Wish I could actually help on this one though.
Last edited by Jared Magneson on Mon Aug 13, 2018 9:59 pm; edited 1 time in total
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Re: Possible Charged Particle Field
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That makes sense to me. You may have noticed I've been thinking of your finding strictly in terms of surface area. In my last post I interpreted your 'emission volume' as an expanding dt surface  "We also know, given an interval of time, a charged particles’ emissions will expand outward from the particle, as on an expanding spherical layer". The entire emission volume would involve all subsequent time differentials which we don't need.
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That makes sense to me. You may have noticed I've been thinking of your finding strictly in terms of surface area. In my last post I interpreted your 'emission volume' as an expanding dt surface  "We also know, given an interval of time, a charged particles’ emissions will expand outward from the particle, as on an expanding spherical layer". The entire emission volume would involve all subsequent time differentials which we don't need.
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LongtimeAirman Admin
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Re: Possible Charged Particle Field
Yeah, I was thinking the same way, only really seeing it as the top layer of the volume, but still didn't put it together that I should be using the surface area instead. Miles had to remind me as I looked over one of his papers. But my method is sound and it was only the initial equation that needed to be fixed, so the general outcome is the same.
It is good to know that you can understand what I have written. That I am making sense. So you are doing a bit more than just cheerleading. I want to show how I find these things, and why the math is the way it is, so others can see what being mechanical means. So they can run with it themselves and maybe find things that everyone else has missed.
It is good to know that you can understand what I have written. That I am making sense. So you are doing a bit more than just cheerleading. I want to show how I find these things, and why the math is the way it is, so others can see what being mechanical means. So they can run with it themselves and maybe find things that everyone else has missed.
Nevyn Admin
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Re: Possible Charged Particle Field
Let's have a look at the Radius function.
The radius function, R, must return the radius of a particle given a time differential, t, where t is in unittime, for example, the number of seconds. When t=0, the current radius is returned. When t>0, a future radius is returned. When t<0, a past radius is returned.
We use the equation of motion s = (at^2)/2 to find the displacement of the radius given the time it occurs over, and the acceleration of the radius. We already know the time, d/c, but we need to know the value of gravity. I have found differing values for the gravity of the proton and need to decipher which one is applicable and why. For now, we will let it stand as the variable a.
Let:
r = current radius of emitter
a = gravity of emitter
t = amount of time the motion occurs over
Then:
If t < 0
R(t) = r  at^2 / 2
Else
R(t) = r + at^2 / 2
Unfortunately, passing in a negative time does not work as well in the math as I wanted it to. The time is squared which means the negative sign is lost so it needs to be extracted from the time or used in a different way to avoid this. I am going to redefine the R function so that it doesn't need to handle that part.
Let R be a function that returns the change in radius given an amount of time, t, where t>=0.
That leaves the subtraction from the current radius out of the R function and we can deal with the minus sign because we know that we are moving backwards in time in this equation.
So R becomes:
R(t) = (at^2)/2
Putting that into D gives us:
D(d) = C( r  R( d/c ) )/d^2
= C( r  (a/2)(d/c)^2 )/d^2
Now we can see a bit more clearly why the relativistic equation reduces to the quantum version. I said earlier that as d becomes very small, the ratio of d:c approaches 0. In the above equation, that causes the change in radius to be so small that it can be thought of as just being the radius. i.e. r  (a/2)0^2 = r.
The values for r and a are treated as constants here, but they really represent the emitter. It might be a proton or it might be a star. For a given problem they may be constants but in others they will need to be specified. So I would rewrite the equation like this:
D( r, a, d ) = C( r  (a/2)(d/c)^2 )/d^2
to reflect that r, a, and d are all parameters. I'm not sure if that is how mathematicians specify such things but I know programmers will understand.
The radius function, R, must return the radius of a particle given a time differential, t, where t is in unittime, for example, the number of seconds. When t=0, the current radius is returned. When t>0, a future radius is returned. When t<0, a past radius is returned.
We use the equation of motion s = (at^2)/2 to find the displacement of the radius given the time it occurs over, and the acceleration of the radius. We already know the time, d/c, but we need to know the value of gravity. I have found differing values for the gravity of the proton and need to decipher which one is applicable and why. For now, we will let it stand as the variable a.
Let:
r = current radius of emitter
a = gravity of emitter
t = amount of time the motion occurs over
Then:
If t < 0
R(t) = r  at^2 / 2
Else
R(t) = r + at^2 / 2
Unfortunately, passing in a negative time does not work as well in the math as I wanted it to. The time is squared which means the negative sign is lost so it needs to be extracted from the time or used in a different way to avoid this. I am going to redefine the R function so that it doesn't need to handle that part.
Let R be a function that returns the change in radius given an amount of time, t, where t>=0.
That leaves the subtraction from the current radius out of the R function and we can deal with the minus sign because we know that we are moving backwards in time in this equation.
So R becomes:
R(t) = (at^2)/2
Putting that into D gives us:
D(d) = C( r  R( d/c ) )/d^2
= C( r  (a/2)(d/c)^2 )/d^2
Now we can see a bit more clearly why the relativistic equation reduces to the quantum version. I said earlier that as d becomes very small, the ratio of d:c approaches 0. In the above equation, that causes the change in radius to be so small that it can be thought of as just being the radius. i.e. r  (a/2)0^2 = r.
The values for r and a are treated as constants here, but they really represent the emitter. It might be a proton or it might be a star. For a given problem they may be constants but in others they will need to be specified. So I would rewrite the equation like this:
D( r, a, d ) = C( r  (a/2)(d/c)^2 )/d^2
to reflect that r, a, and d are all parameters. I'm not sure if that is how mathematicians specify such things but I know programmers will understand.
Nevyn Admin
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Re: Possible Charged Particle Field
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After rereading a couple of Miles’ third wave and part of his long Mercury precession papers earlier today, I’d say you’re using the motion equation s = (at^2)/2, to find the displacement of the radius exactly as Miles has done.
Critically speaking, that’s all I’ve got.
You’re making charge field history. What next?
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I’m sure mathematicians and programmers would agree, D( r, a, d ) appears radical. The charge density received from a charged particle depends on the particle’s radius, gravity and distance. Standard math doesn’t list all the parameters used on the right side of an identity equation as an index, or whatever that is, on the left. I would say a  the gravity of the emitter  is the only emitter “constant”. It’s difficult to tell the difference between distances and an acceleration without accompanying units  meters, seconds or sec^2.So I would rewrite the equation like this:
D( r, a, d ) = C( r  (a/2)(d/c)^2 )/d^2
to reflect that r, a, and d are all parameters. I'm not sure if that is how mathematicians specify such things but I know programmers will understand.
After rereading a couple of Miles’ third wave and part of his long Mercury precession papers earlier today, I’d say you’re using the motion equation s = (at^2)/2, to find the displacement of the radius exactly as Miles has done.
Critically speaking, that’s all I’ve got.
You’re making charge field history. What next?
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LongtimeAirman Admin
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Re: Possible Charged Particle Field
Yeah, I thought it would be nonconformant, but I wanted to express that they may be variables in some circumstances while they may be constants in others. This is especially true when dealing with stars, planets and moons. Every entity will have its own radius and gravity value, whereas if we are looking at protons, then they all share the same values for those. But electrons and neutrons will be different, so it isn't just a micro vs macro type of thing. Even for our purposes in this app, we will need to handle different values once we bring electrons into the mix. We could probably use the same values for protons and neutrons though.
What's next?
Seriously though, I am working on the Charge Density function, C, to figure out if charge density can be calculated in the way that I want it to be. I want an expression that gives us the charge density, but only for a unitarea of the emitter, that can then be expressed on the receiver.
Question: What do you get if you take the volume of a sphere and divide it by the surface area of that sphere?
Conceptually, you should get the volume per unitarea.
A Dimensional Analysis tells you that the value is a meter (assuming you are using meters as your length units). It is 1 dimensional, not 3, so how does it represent a volume? Because it is a radius!
If you take the equation for the volume of a sphere (4/3 * PI * r^3) and divide it by the equation for the surface area of a sphere (4 * PI * r^2), you get r/3. Therefore, if my interpretation is correct, the volume per unitarea is equal to the volume of a sphere with a radius that is 1/3 of the original sphere's radius.
Does that sound reasonable?
Also, gravity. With the equation s = at^2/2, we can find the change in radius given an amount of time (which will be the time between frames but also include any time discrepancy as a result of distance separation) to find the amount of space that the other particles have taken up as a result of their expansion. We convert that into a vector that pulls our particle in that direction, by that amount. We do that for every other particle and sum them all together to get a resultant gravity vector. I'm not sure this is correct, since the other particles will expand regardless of what our current particle is doing. But in this math, 2 particles on either side of our particle will result in no motion. It doesn't feel quite right, but I can't see another way to do it, yet.
What's next?
Brain, in every episode of Pinkie and the Brain, wrote:The same thing we do every night, Pinkie, plan to take over the world!
Seriously though, I am working on the Charge Density function, C, to figure out if charge density can be calculated in the way that I want it to be. I want an expression that gives us the charge density, but only for a unitarea of the emitter, that can then be expressed on the receiver.
Question: What do you get if you take the volume of a sphere and divide it by the surface area of that sphere?
Conceptually, you should get the volume per unitarea.
A Dimensional Analysis tells you that the value is a meter (assuming you are using meters as your length units). It is 1 dimensional, not 3, so how does it represent a volume? Because it is a radius!
If you take the equation for the volume of a sphere (4/3 * PI * r^3) and divide it by the equation for the surface area of a sphere (4 * PI * r^2), you get r/3. Therefore, if my interpretation is correct, the volume per unitarea is equal to the volume of a sphere with a radius that is 1/3 of the original sphere's radius.
Does that sound reasonable?
Also, gravity. With the equation s = at^2/2, we can find the change in radius given an amount of time (which will be the time between frames but also include any time discrepancy as a result of distance separation) to find the amount of space that the other particles have taken up as a result of their expansion. We convert that into a vector that pulls our particle in that direction, by that amount. We do that for every other particle and sum them all together to get a resultant gravity vector. I'm not sure this is correct, since the other particles will expand regardless of what our current particle is doing. But in this math, 2 particles on either side of our particle will result in no motion. It doesn't feel quite right, but I can't see another way to do it, yet.
Nevyn Admin
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Re: Possible Charged Particle Field
Brain, in every episode of Pinkie and the Brain, wrote:
The same thing we do every night, Pinkie, plan to take over the world!
Lol!!!
Cr6 Admin
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Re: Possible Charged Particle Field
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Pinky and the Brain are about to implement gravity.
Curious about the outcome? See: animaniacs pinky and the brain
https://gifer.com/en/oaf
////\\\\////\\\\////\\\\////\\\\////
I thought we were going to implement gravity by redrawing expanding particles. The screen dimensions would need to be recalculated along with the acceleration of the expanding particles.
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Pinky and the Brain are about to implement gravity.
Curious about the outcome? See: animaniacs pinky and the brain
https://gifer.com/en/oaf
////\\\\////\\\\////\\\\////\\\\////
I thought we were going to implement gravity by redrawing expanding particles. The screen dimensions would need to be recalculated along with the acceleration of the expanding particles.
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LongtimeAirman Admin
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Join date : 20140810
Re: Possible Charged Particle Field
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Sorry Nevyn, I don't understand your resultant gravity vector idea. If particles are outside each other's charge range will gravity still be calculated? Real expanding particles seems easier.
Concerning the reasonableness of the notion of a volume of a sphere divided by the sphere’s surface area. You’ve shown how they – the surface area and volume  are related. A “volume of particle emission” becomes an expanding spherical layer, a surface area with a differential time thickness dt. The further an object is, the smaller the amount of the expanding surface area the receiver will intercept. Sounds perfectly reasonable to me.
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Sorry Nevyn, I don't understand your resultant gravity vector idea. If particles are outside each other's charge range will gravity still be calculated? Real expanding particles seems easier.
Concerning the reasonableness of the notion of a volume of a sphere divided by the sphere’s surface area. You’ve shown how they – the surface area and volume  are related. A “volume of particle emission” becomes an expanding spherical layer, a surface area with a differential time thickness dt. The further an object is, the smaller the amount of the expanding surface area the receiver will intercept. Sounds perfectly reasonable to me.
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LongtimeAirman Admin
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Re: Possible Charged Particle Field
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Area and Volume of a Circle and Sphere Using Simple Geometry
http://physicsinsights.org/spherevolume1.html
I don't see a problem. Check your thinking against this site.Therefore, if my interpretation is correct, the volume per unitarea is equal to the volume of a sphere with a radius that is 1/3 of the original sphere's radius.
Area and Volume of a Circle and Sphere Using Simple Geometry
http://physicsinsights.org/spherevolume1.html
.(14) Volume(sphere) = (1/3)*r*A
where A now represents the surface area of the sphere. ...
(15)The Surface(sphere) = 4Pi*r^2
LongtimeAirman Admin
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Re: Possible Charged Particle Field
LongtimeAirman wrote:
I thought we were going to implement gravity by redrawing expanding particles. The screen dimensions would need to be recalculated along with the acceleration of the expanding particles.
No, we can't do that for at least 2 reasons. The first one mathematical and the second physical.
1) The numbers explode!
We are working within a certain precision which is set by the computer that the application is running on and the language used to implement it. In our case, we have 64bit floating point numbers in Javascript, but on a GPU, such as in a shader, we may only have 32bit floating point values. On a phone, we might only have 16bit numbers. No matter how large that is, it is still finite. We could be working with 1024bit numbers and it still would not be enough. If you just keep increasing the radii, then you will eventually, actually rather quickly, reach the upper limit of what your numbers can store.
There are ways around this limit. You can work with arbitrarily sized numbers that do not have an upper limit, except for the amount of memory in your system. However, they introduce a speed limit. The larger your numbers are, the slower they are to use in any calculation. Eventually, again rather quickly, the app grinds to a halt just trying to add 2 numbers together.
2) It doesn't represent the actual physics.
A 3D system is a Newtonian system, rather than a Relativistic one. The 3D engine will render all geometry as it is now, no matter how far it is from the camera. It does take distance into account to make things smaller the further they are from the camera, but it does not take the time discrepancy that that distance represents into account. We can make it do that, but only with considerable effort, and it is not as easy as it first looks.
You see, it isn't just the size that we see in the past, it is also the position (and velocity, spin, etc). When we look at the Sun from the Earth, we are seeing it at a size that it was 8 minutes ago, but we are also seeing it in the position that it was in 8 minutes ago. This is also a problem for charge calculations. My equation above takes care of the time discrepancy by using a smaller radius to calculate the initial emission, but it does not take the position difference into account. In its current state, it can't do that. In fact, it might not be feasible to do that because while that charge was travelling from the source to the receiver, the emitter could have had a collision and changed trajectory. We have no way of knowing that.
So we are left to find another way. It isn't going to be perfect, but we may be able to get pretty close and hopefully minimize the differences.
My idea is to treat everything as if it is always at the same size, but use adjusted values from all other particles that the current particle is interacting with. You always need a reference point, which is the current particle for us, so that you can determine how things look from that point. They may look completely different from a different point, but that is irrelevant to our current particle, for it is not over there, it is here.
Nevyn Admin
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Re: Possible Charged Particle Field
LongtimeAirman wrote:Sorry Nevyn, I don't understand your resultant gravity vector idea. If particles are outside each other's charge range will gravity still be calculated? Real expanding particles seems easier.
I'm not sure about that. There is some distance where the change becomes so small as to be irrelevant. We can choose to find that distance and ignore everything outside of it just like we are with charge. Or we might just add it in anyway. It will probably be just as much computational effort to determine if it is too far away as it is to just calculate and add it in, which is not the case with charge as that is expensive to calculate.
LongtimeAirman wrote:Concerning the reasonableness of the notion of a volume of a sphere divided by the sphere’s surface area. You’ve shown how they – the surface area and volume  are related. A “volume of particle emission” becomes an expanding spherical layer, a surface area with a differential time thickness dt. The further an object is, the smaller the amount of the expanding surface area the receiver will intercept. Sounds perfectly reasonable to me.
Not quite correct. Or it is in some ways, but maybe a bit misleading. The further an emitter is from the receiver, the larger the surface area is at the location of the receiver. That is, if we calculate a unitarea on the emitter at the time of emission, then that unitarea expands as that emission travels away from the emitter. That is what the outer 1/d^2 is doing in my equation. It is finding the amount of emission over the same unitarea on the receiver and the d^2 reduces the charge density to reflect the same unitarea at the receiver. The unitarea stays the same and we let the charge density change but we could just as easily find the new area that the emission covers at the receiver.
In fact, I think we should do that. We should let the emission travel out uninhibited until it reaches the receiver. We then calculate the size of the original unitarea at that distance. Then we divide that, let's call it the receivedarea, by the area that it interacts with on the receiver. This creates an equation that calculates the interaction of one area, on the emitter, and another area, on the receiver. Therefore, it can handle the difference in area that the receiver has with respect to the emitter. I might investigate that at some point. It may just lead to the same 1/d^2 relationship, but it is worth finding out.
LongtimeAirman wrote:I don't see a problem. Check your thinking against this site.
Area and Volume of a Circle and Sphere Using Simple Geometry
http://physicsinsights.org/spherevolume1.html
I knew the math was correct, not because I searched for it, but because I plugged some values into it and saw the relationship directly. What I am unsure of is my interpretation of it being the radius of a sphere to represent a volume. That site does not explicitly state it, but I was using the same concepts as they are (as in a pyramidlike shape from the surface to the center).
What I am trying to get at with that analysis, is that it is the whole volume that is emitting, not just the surface, even though it only emits at the surface. How can that be? Easy, because not all of that emission was created (for want of a better term) at the same time, but it is all emitted from the surface at the same time.
This is easier to see with stars and planets, than protons or small particles. I am actually wondering if there is a different equation for particles as compared to composite entities like stars. Anyway, with a star or planet, we have a body made up of smaller particles and it is each of those particles that is actually emitting. The sum of that emission becomes the emission of the body. However, the emission that came from the very center of that body is older than the emission that came from the surface particles. But age doesn't really matter. It is the sum of emission at the surface that we care about because that is the amount of emission that will travel away from the emitter and eventually collide with a receiver.
That gets us the total charge emission from the volume (with an assumption that the whole sphere is emitting, which we know is incorrect) at the surface. I then divided that by the surface area to get to a unitvolume or the amount of volume that is behind a unitarea. That should represent the amount of emission that is moving towards the receiver, with some margin of error. What we really should do, is determine the surface area of reception, translate that into an area of emission and use that to calculate the amount of emission. I may investigate that as well.
Nevyn Admin
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Re: Possible Charged Particle Field
I had an idea a few months ago that might be able to help us here. In my posts above, I mentioned that there is a problem with reaching back into the past because you can only use the current data to do so. This ignores any interactions between the time of emission and reception that might have occurred to the emitter because there is no way to determine that. The best we could do is assume a reversal of the current velocity, spin, etc, which is not really good enough. It might work okay with stars and planets, but not at all with particles. Although, particles have an advantage in that they are so close to each other that we aren't looking very far into the past, but they also collide a lot more often and the results of those collisions are more drastic if compared to a planet.
The idea I had was to look at everything as an event. Each event has a time that it occurred, the position and orientation it was in, the resultant velocity and spin, etc. Essentially, it is the state of the entity at that event, or just after that event really. From there, we can determine its current position and orientation by using the time since that event and apply it to the velocity and spin. So we never change the position of an entity, for example, we just calculate its current one. Only a collision, which is the event, can cause those values to change.
Now, we can extend that a bit to keep a certain amount of history. We store the last 10 events, for example, so we can actually determine the complete story of that entity at least that far back into the past. We have a path, not just a line. Jared will be familiar with this because it is the basis of keyframes that he works with. However, he is using it to drive the complete animation, whereas I just want a window of it to represent the past.
The idea I had was to look at everything as an event. Each event has a time that it occurred, the position and orientation it was in, the resultant velocity and spin, etc. Essentially, it is the state of the entity at that event, or just after that event really. From there, we can determine its current position and orientation by using the time since that event and apply it to the velocity and spin. So we never change the position of an entity, for example, we just calculate its current one. Only a collision, which is the event, can cause those values to change.
Now, we can extend that a bit to keep a certain amount of history. We store the last 10 events, for example, so we can actually determine the complete story of that entity at least that far back into the past. We have a path, not just a line. Jared will be familiar with this because it is the basis of keyframes that he works with. However, he is using it to drive the complete animation, whereas I just want a window of it to represent the past.
Nevyn Admin
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Re: Possible Charged Particle Field
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You're covering new ground, I may not be the only one with a confusion or two. Must have pictures. Feel free to mark or make corrections or suggestions.
The emitter is on the left, and the receiver is on the right. You’ve recently developed a relativistic charge field, based on expansion theory, where charge emitted  C(), the particle radius  R() and the amount of charge received  D() allow us to model charged particles of almost any size, quantum or stellar.
Currently, the charge density received calculation is done at the receiver’s N/S,E/W,F/B points. I believe that this method doesn’t seem capable of differentiating between a relativistic case or not, so you are coming up with a better method.
I'm not sure I understand your r/3 spherical volume. I would say that the volume of the sphere that is the source for all emission felt by the receiver is the small conic section of the emitter, the base of the conic section is pointed toward the receiver. We can see the expansion of the surface area intercepted by the receiver.
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You're covering new ground, I may not be the only one with a confusion or two. Must have pictures. Feel free to mark or make corrections or suggestions.
The emitter is on the left, and the receiver is on the right. You’ve recently developed a relativistic charge field, based on expansion theory, where charge emitted  C(), the particle radius  R() and the amount of charge received  D() allow us to model charged particles of almost any size, quantum or stellar.
Currently, the charge density received calculation is done at the receiver’s N/S,E/W,F/B points. I believe that this method doesn’t seem capable of differentiating between a relativistic case or not, so you are coming up with a better method.
I'm not sure I understand your r/3 spherical volume. I would say that the volume of the sphere that is the source for all emission felt by the receiver is the small conic section of the emitter, the base of the conic section is pointed toward the receiver. We can see the expansion of the surface area intercepted by the receiver.
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LongtimeAirman Admin
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Re: Possible Charged Particle Field
Nevyn wrote:Jared will be familiar with this because it is the basis of keyframes that he works with. However, he is using it to drive the complete animation, whereas I just want a window of it to represent the past.
Indeed, and ideally my simulations would (and will) borrow as much from yours as possible. I just don't know how to get it into Maya yet really! Obviously keyframes and editing are useful for a videostyle presentation, but the simulation itself needs to be on or approach this level for me at some point. I'm following along pretty well, though you still work some serious magic to me. And really appreciate what you guys are doing here.
Jared Magneson Posts : 514
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Re: Possible Charged Particle Field
Sorry for the confusion, I'm just playing with ideas. Seeing what's possible. Some of those ideas are not for this app, but are in the same area. There isn't one magic equation for everything. There are little subtleties that can change the equations you want to use in a given problem or you might want a different kind of information so you go about it in a different way. I am not being clear about that at all. I'm just dumping my thoughts at this stage and seeing if they can stand up by themselves. Not very fair to anyone trying to understand me!
The use of charge points (N,S,E,W,F,B, for example) is still viable and the way I want to go. I'm quite proud of that little algorithm! It has no relation to being relativistic because the charge points are about applying the charge to the receiver, whereas being relativistic is about supplying charge by the emitter. It is the emission that happened in the past maybe because it is a long way away. The reception of that charge is a local event. It happens now, so no need for Relativity on that side of the interaction. If you want to know if something is, or needs to be, relativistic, just look for time discrepancies. They can be masked as distances.
Yes, it is the conic section that I was trying to get at. We want to know how much volume is in that section as that is the amount of charge that the receiver has to deal with. If we imagine that where that conic section touches the surface is a unitarea, then that whole conic section is a unitvolume. If my interpretation is correct, then that section has a volume that is equal to the volume of a sphere with 1/3rd of the radius of this sphere. So imagine that we take the sphere, create another sphere that has 1/3rd the radius, and then squash that smaller sphere into the shape of the conic section. It should match. That's all I'm trying to say about that. It seems logical to me, but am open to criticism as well.
The use of charge points (N,S,E,W,F,B, for example) is still viable and the way I want to go. I'm quite proud of that little algorithm! It has no relation to being relativistic because the charge points are about applying the charge to the receiver, whereas being relativistic is about supplying charge by the emitter. It is the emission that happened in the past maybe because it is a long way away. The reception of that charge is a local event. It happens now, so no need for Relativity on that side of the interaction. If you want to know if something is, or needs to be, relativistic, just look for time discrepancies. They can be masked as distances.
Yes, it is the conic section that I was trying to get at. We want to know how much volume is in that section as that is the amount of charge that the receiver has to deal with. If we imagine that where that conic section touches the surface is a unitarea, then that whole conic section is a unitvolume. If my interpretation is correct, then that section has a volume that is equal to the volume of a sphere with 1/3rd of the radius of this sphere. So imagine that we take the sphere, create another sphere that has 1/3rd the radius, and then squash that smaller sphere into the shape of the conic section. It should match. That's all I'm trying to say about that. It seems logical to me, but am open to criticism as well.
Nevyn Admin
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Re: Possible Charged Particle Field
Another way your picture works for me is that we can work backwards. We look at the emission radius at the receiver and project that back onto the emitter to find the actual area required by this interaction. This method does not work with my charge points because we are trying to find the charge density across the entire receiver, not at particular points on it. You could use that for a star pushing out on a planet when you just want the amount of repulsion and don't care about spins.
Nevyn Admin
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Re: Possible Charged Particle Field
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Maybe the charge point (N,S,E,W,F,B) method can be used for the relative case. You mentioned tracking state changes. Why not track the last 10 (or however many) state changes  and when they occurred  for each particle’s N,S,E,W,F,B cardinal locations. The past position of the emitter may then be compared to the present in order to arrive at a gravity component to the velocity or spin changes based on the change in charge received from that particle in the past.
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Maybe the charge point (N,S,E,W,F,B) method can be used for the relative case. You mentioned tracking state changes. Why not track the last 10 (or however many) state changes  and when they occurred  for each particle’s N,S,E,W,F,B cardinal locations. The past position of the emitter may then be compared to the present in order to arrive at a gravity component to the velocity or spin changes based on the change in charge received from that particle in the past.
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LongtimeAirman Admin
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Re: Possible Charged Particle Field
That is actually a reason why we can't use that method. There are too many charge interactions so it would require too much history. At 60 frames per second we would have to record 60 state changes only to have 1s worth of history because charge interactions tend to happen over many frames.
Even if we could use it though, we wouldn't need to store history for each charge point since they are just used to find the resultant velocity and spin, which is what we would store. The charge points are just a way to break up the surface that is being bombarded with charge so that we can find small differences and apply them individually. They still collectively represent the surface of the receiver and they are always put back into a single vector for velocity and a single quaternion for spin.
Maybe we could tackle it slightly differently and still use the history. Instead of recording every collision, we just sample every entity at regularly intervals and record its state. The recorded state becomes the current state minus the previous state. This makes it easier to determine how much history you will store too. Because we set the time between samples, we know the sample rate. Multiply that by the number of samples and you have the time that those samples cover.
So if we sampled every second and kept 60 state changes, then we would have 1 minute of approximate history.
However, I don't think that is to be used in this app. We are dealing with protons not planets and stars, so we don't really have a history problem.
Even if we could use it though, we wouldn't need to store history for each charge point since they are just used to find the resultant velocity and spin, which is what we would store. The charge points are just a way to break up the surface that is being bombarded with charge so that we can find small differences and apply them individually. They still collectively represent the surface of the receiver and they are always put back into a single vector for velocity and a single quaternion for spin.
Maybe we could tackle it slightly differently and still use the history. Instead of recording every collision, we just sample every entity at regularly intervals and record its state. The recorded state becomes the current state minus the previous state. This makes it easier to determine how much history you will store too. Because we set the time between samples, we know the sample rate. Multiply that by the number of samples and you have the time that those samples cover.
So if we sampled every second and kept 60 state changes, then we would have 1 minute of approximate history.
However, I don't think that is to be used in this app. We are dealing with protons not planets and stars, so we don't really have a history problem.
Nevyn Admin
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Re: Possible Charged Particle Field
Okay, this is really fun and only slightly offtopic.
But if you load up the "Proton > Stack 3" sim and stare at it for a bit, then look away somewhere else, your vision keep warping for a moment! Reality appears to ripple, the optics are fun. Just thought it was cool!
But if you load up the "Proton > Stack 3" sim and stare at it for a bit, then look away somewhere else, your vision keep warping for a moment! Reality appears to ripple, the optics are fun. Just thought it was cool!
Jared Magneson Posts : 514
Join date : 20161011
Re: Possible Charged Particle Field
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Good longitude to you.
Jarred, when I stare at things like that I see auras  of course I 'm sure that's due to strictly visual persistence. On the other hand I do like thinking of auras. At first I thought it was New Age nonsense. Now, I merely find it hard to believe anyone could actually perceive our real charge field auras.
Nevyn, I must ask, using your examples, I know how to set a particle's: type; position; velocity; and orientation. I believe the only thing missing  I've made several attempts  how should I set an initial spin?
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Good longitude to you.
Jarred, when I stare at things like that I see auras  of course I 'm sure that's due to strictly visual persistence. On the other hand I do like thinking of auras. At first I thought it was New Age nonsense. Now, I merely find it hard to believe anyone could actually perceive our real charge field auras.
Nevyn, I must ask, using your examples, I know how to set a particle's: type; position; velocity; and orientation. I believe the only thing missing  I've made several attempts  how should I set an initial spin?
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LongtimeAirman Admin
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