Why do fusors work?

[ab0032]

The question running around my head is, why do fusors work? From evidence by many people here on the forum who have done it, we know they do work, and not only for a brief moment but apparently in a steady state.

If a fusor runs for a while, I would think that the ions as well as any electrons would thermalize, they would loose energy by all the radiation going out. I would believe that slower ions would go into smaller orbits around the grid and form a positive cloud that would stop the full power of the electric field to reach out far. What keeps the ions speed high enough for fusion?

Do slower ions hit the grid and become neutralized, enabling the atoms to escape again, being replaced by new ions created with higher energy?

Is star mode, which is mentioned off and on a necessary requirement for a working fusor? Have there been fusors that produced neutrons without star mode? What explanations are there for star mode? What happens in star mode?

Cheers,
Alex

[Chris Bradley]

Alexander Biersack wrote:
> The question running around my head is, why do fusors work? From evidence by many people here on the forum who have done it, we know they do work, and not only for a brief moment but apparently in a steady state.
> If a fusor runs for a while, I would think that the ions as well as any electrons would thermalize, they would loose energy by all the radiation going out. I would believe that slower ions would go into smaller orbits around the grid and form a positive cloud that would stop the full power of the electric field to reach out far. What keeps the ions speed high enough for fusion?
> Do slower ions hit the grid and become neutralized, enabling the atoms to escape again, being replaced by new ions created with higher energy?

They do, indeed, thermalise. They thermalise almost instantly and, in statistical terms, *all* the ions and electrons(as in, 99.9999%) simply thermalise immediately. It is just pure, brutal, brute force of accelerating 'suicidal' ions, of which a near-zero number get into fusion collisions.

It's pretty much the same question as 'why does anyone win the lottery?'.

A fusor works by accelerating billions *of* billions ions per seconds, most of which just go to generate heat in the fusor. A few million get to fuse. It's just flat out luck (or un-luck, depeding on whether you wanted more than a few millions fusions!!).

There are a few threads about this, if you dig deep. Here's one you can add to, as it didn't seem to go very far!:

download_thread.php?site=fusor&bn=fusor ... 1278342911

[Richard Hull]

Chris got it right on. We "bullhead" the operation. The sheer simplicity is the reason the fusor will never, and can never, produce one tenth of a watt of output power or even one ten thousandth of one watt of net power.

The above and Chris' explanation is why the fusor, as we make it, will not lend itself to be modeled mathematically. I would imagine that the best shot at modeling of a fusor would be to take a large integrated and averaged cross section of accurate operational results and model backwards from those into a workable mathematical fabric, provided someone was so inclined and so blessed with mathematical prowess and a grasp of all the physics going on.

With the fusor, it is not about making energy it is all about wasting it to get a desired result; i.e.... Space heater warms a room. We throw away valuable electrical energy...object?... to keep the room warm and habitable; a laudable and often highly desirable reward for the massive loss of energy. In our case, we throw away all that energy to be able to say, with confidence, that we have done nuclear fusion....or.... to obtain neutrons for experiments. We just throw a quadrillion or so pairs of dice and only count how many "snake eyes" or "box cars" we get.

Star mode is not a must see to be doing fusion, but in most all successful fusors it is always seen, but often fades a bit at "sweet spot" during intense operation. Conversely, star mode is easily and often seen in fusors doing absolutely zero fusion. As you can see, star mode cannot, in and of itself, act as a visusal herald that fusion is taking place.

Richard Hull

[David Geer]

Star mode isn't necessary but is a good visual for seeing imperfections in grid structure and magnetic fields. If you had a strong enough magnetic field coming from the chamber shell, you'd see the vented particles bend and fall back into the plasma core.

Everyone will agree though, that it is a beautiful phenomena in the course of experimentation.

[Dan Tibbets]

Da&#@&%
I had written a long rambling post, then managed some keystroke combination that erased it.
A more condensed effort. Everything said is true, though by greatly increasing the size of a gridded fusor, you could get up to perhaps a few watts of fusion power. Of course to do so you would need millions of dollars and a large warehouse to build it in, and it would take millions of watts of input power to operate. Breakeven can never even be approached .

Coulomb collisions and thermalization processes are not instantaneous, it may take up to a few microseconds to occur depending on density and energy/ temperature. At low energy the coulomb crossection is huge, but it decreases at a rate of ~ T^1.75 power. The fusion crossection for deuterium increases, but at a slower rate once you pass a few hundred KeV. There is a sweat spot where you can minimize the Coulomb scattering crossection to fusion crossection ratio. For D-T this is at about 80 KeV. The coulomb crossection is only ~ 10 times the D-T fusion crossection. For D-D you might reach similar ratios at ~ 1,000 KeV, but at this point you loosing ground fast due to other considerations like bremsstrulung, and input energy to reach those energies. Between 100 and 1,000 KeV the fusion crossection is increasing at a rate of ~ T^0.75 power(rough visual guess from looking at the graphs). But the input energy increase is T^1 power. IE: the fusion rate may increase ~ 7 fold, but the input energy increased 10 fold. You are losing ground, unless you have very efficient energy recovery mechanisms. Steam would allow you to only reach breakeven in this situation, and that is at much higher cost and difficulty.

The MFP (Mean Free Path) is defined as a collision that results in a 90 degree deflection. smaller angle collisions are more frequent but the net effect is the same. I have heard comments that with D-D fusion with energies of ~ 60+KeV the Coulomb collision thermalization issues become trivial. I doubt that is quite the case, the angular momentum scattering should only take one MFP or at most two collisions to fully thermalize the angular momentum, depending on the geometry. But, perhaps more important is the upscattering and down scattering collisions. One estimate that I did by watching one of Falsteed's (sp?) simulations was that it took ~ 3-6 collisions before a Maxwell thermalized distribution was reached when starting from a monoenergetic population. This would still result the fuel ions approaching a near Maxwellian distribution, but perhaps there would still be a significant fuel burn up in this interval , and the dynamics would always be better (a tiny to moderate amount) than starting with a thermalized plasma).

Also, keep in mind that thermalization is not always bad. In the Polywell, thermalization under localized conditions may actually be very helpful. And Tokamaks may work despite global thermalization. The important point here is that the fusor is limited ultimatly by the physical grids. The ions can only make ~ 10-20 passes before the grid is hit and all the work done to accelerate the ion is lost as heat. The walls have a similar effect in a thermalized plasma due to upscattering. If any electrostatic scheme can reach breakeven it will probably be something similar to the Polywell, where tricks are played with grid transparency, thermalization management and density boosting.

In a breakeven + fusion reactor the ion has to survive within the reacting volume for a certain amount of time that is dependant on the temperature dependent fusion crossection (complicated by thermalization issues) and the density squared.
If you generate ~ 10 MeV per fusion but you have to provide 1,001 10 KeV fuel ions / fusion, you will fall short (unless you can recover some of the otherwise wasted heat energy- then it is more forgiving, except you have to also consider the conversion efficiency of the fusion energy). Because of this the residing time that the ion must achieve for breakeven is easily calculated or compared.

In a typical fusor the ion (or energetic neutral) residing time may be ~ 20 microseconds at a density ~ 10^20 particles / M^3

In a Tokamak the residing time may be ~ 800 seconds at the same density.

In a Polywell (if it works) the residence time may be ~ 20 milliseconds or more with a density of ~ 10^22/ M^3.

Note: that the residence times for the Polywell (assuming it is operating in a thermalized mode) and the Tokamak compared to the density results in a similar result, or triple product (Time * density^2 * Temperature) and it shows how far behind the gridded fusor is.

Finally, a central concentration of positively charged ions can occur and has been observed. This is generally referred to as a virtual anode. It's magnitude is usually small compared to the cathode provided potential well due to inefficiencies, scattering limiting the efficiency of the ion convergence in a spherical geometry. Theoretically it could become a problem if everything else is operating exceptionally well. In the Polywell, I think Bussard set a limit of ~ 20% on the maximum virtual anode strength that would be desirable for several reasons.

PS: Aren't you glad I had to rewrite a shorter version of my thoughts

Dan Tibbets

[Chris Bradley]

Dan DT wrote:
> by greatly increasing the size of a gridded fusor, you could get up to perhaps a few watts of fusion power.

How do you arrive at this conclusion?

A fusor is, *at best*, a convergent device wherein ions are kept a little longer recirculating than just plain hammering them into a target. If that is so, increasing its size won't increase the convergence. It is still 4pi.steradians worth of convergence, whatever its size. Why would size make a difference? I'm not saying it doesn't, but I do not see that it is trivially obvious.

[Actually, in regards efficiency, I am beginning to suspect small size and high ratio grid:shell dimensions are better, because the fields are stronger, the ions spent proportionately longer at the fastest speeds, and non-neutral conditions may be more prevalent (thus more ions).]

[Carl Willis]

This thread was perhaps destined from the beginning, like so many others here, to end up deep in the weedy outfield of pure and far-ranging speculation, questionable unsourced figures, and a discursive free-for-all.

Looking at Alexander's original question, I think he was specifically asking why fusion occurs at a steady-state rate despite some hypothetical mechanisms he postulates would terminate the process quickly. So the question seems to revolve around the sources and sinks of ions in steady operation. The thread's title isn't usefully specific, but I think the original question was specific enough to usefully focus the replies.

Thanks
Carl

[Dan Tibbets]

To answer Chris Bradley, I'm under the impression that there is not much central focus in a gridded fusor, and my impression is that you think there is even less- at least in terms of centrally concentrated fusor reactions. Lacking this high grate focus or convergence, the fusion rate should scale at ~ the radius ^3- the volume. The input costs should scale at ~ the same rate, so from absolute fusion output increasing the radius 10 fold should increase the fusion rate 1,000 times. The problem is that the input costs also increase ~ 1000 fold.

As Carl Willis points out, the original questioner may not appreciate the tangents. At root levels, comparing a fusor to a beam target machine is the most basic approach and sometimes quite accurate. You accelerate an ion with an applied voltage (this is very easy to do, even old cathode ray tube televisions imparts enough energy for this- at least for a few fusion reactions). This fast ion hits an atomic nucleus square on in a solid target material and fuses if it can overcome the coulomb barrior. I believe this basic approach is how most of the data about nuclear reactions and crossections was accumulated since at least the 1940s. Most of the ions do not hit and result in a fusion, but enough do to collect a lot of information. A gridded fusor does the same- the fusion may occur when the ion hits a target atom on the wire grid or on the vacuum vessel wall. In this case it is exactly like a beam target fusion. The only real difference is that the ion recirculates if it doesn't hit a solid surface and may have 10-20 additional chances to hit another floating neutral atom or ion that is not on a surface. From a very simplistic approach this implies that a fusor shoud be ~ 10-20 times as efficient as the standard beam target approaches used for decades in labs. The difference between this and profitable fusion (excess energy) is a whole different issue. To even begin to think about profitable fusion you have to manage to increase the recirculation many thousands of times, while avoiding putting in additional energy. The target nucleus whether fixed or floating is a very small target and just right collisions occurs very rarely, but with billions upon billions of ions involved, statistically these collisions happen frequently enough to be measurable.

Dan Tibbets

Dan Tibbets

[Steven Sesselmann]

Alex,

I am impressed, you start off by asking a most fundamental question, because nothing in our Universe does anything unless it wants to, you never see water run uphill do you?

So whatever the mysterious conditions are inside a fusor chamber, they make deuterons want to fuse.

Some will disagree with my views, as I do not believe the story that fusion is caused by particles colliding at a high enough energy to overcome the coulomb forces, it may look like this, to the observer, but it's not what causes particles to fuse.

Particles fuse, when the conditions are set up, so that fusing is the better option, in other words, by fusing, the particles can fall to a lower potential. Call it quantum tunneling or whatever you like, it's just nature doing it's thing.

The arrow of time points towards a lower potential , and all matter is ultimately heading that way, what prevents matter from instantly collapsing in on itself, is the low permittivity of free space. A very large amount of energy, surrounded by an almost perfect insulator (fortunately for us).

Keep thinking about this, and suddenly you will realize why fusors work, and more importantly why they don't

Steven

http://www.beeresearch.com.au

[ab0032]

You could still model things in a computer, but not down to the path of all ions and electrons involved. But it can still make sense to model something to get a better understanding of what is going on.

[ab0032]

> So the question seems to revolve around the sources and sinks of ions in steady operation.

Yes, where do the ions come from? Where are they created, where and how are they lost. How does the plasma keep enough energy and not radiate out all its energy?

Of course all the other replies were interesting too, this is more what I want know.

Anything hitting the grid will pollute the plasma. Some ions might be created anywhere half way between the ends of the field, so they have only a share of the full energy.

On the other hand, if things thermalize, we could also see some upscatter, some ions gaining energy, enough to produce fusion.

[ab0032]

Fusion crosssections and beam targets...

As a sceptic I have already doubted the cross section measurements. How where they obtained? I have not found one source, where conditions where specified exactly. How do we know there where no knock on collisions? What densities of ions were used in measurements?

[Dan Tibbets]

It is a fundamental property that items with less potential energy will release energy and visa verse. But to think that this drives fusion is simplistic. If this was the case protons and neutrons in various combinations would very quickly build to Nickel 62, which has the lowest potential energy of any nucleus. The universe- or more specifically, stars would all end up as nickel (or Fe56). But this is not the case. Only those stars massive enough will reach this endpoint. The sun will only reach carbon/ oxygen. Why- because the obtainable temperature is not high enough. Why is that- because of the coulomb barrier. When the products have lower potential energy than the reactants, then excess heat will occur (KE), but this by itself does not drive the speed of a reaction. You can argue that due to statistics and quantum effects a lower potential energy state will eventually happen. But it says nothing about how fast this will occur. Is it 1 second, one year, 100 billion years? This is dependent on the intermediate energy barrier and the availability of applied KE to approach or exceed this barrier.

The same reasoning can be applied to chemical systems. Why does a mixture of Hydrogen and oxygen gas not spontaneously and rapidly burn to water. It is because a threshold has to be exceeded. This is often referred to as Gibbs free energy. There is an intermediate stage that has more potential energy (you have to put in KE) than either the reactants or products. Catalysts work by decreasing this intermediate PE state. And, it is very fortunate that this characteristic of reactions exist, otherwise we would not be here.

In nuclear fusion this intermediate state is due to the tremendously increasing repulsion as protons approach each other. If the KE is great enough (often helped by quantum uncertainty) the protons can approach to ridiculously close distance before the very short range Strong force can overcome this repulsion (and it can only do so to a limited extent. That is why an intermediate nucleus like Nickel has thew lowest PE and Nuclei with more than a few hundred nucleons fall apart. Uranium 235 falls apart into products with less PE with a fairly long half life. Why does it not do so instantly?. Because there is an intermediate state with higher PE that has to be overcome by chance accumulation of additional PE by the nucleus- the extreme high energy tail of a nucleus equivalent of a thermalized gas.
A counter argument might be that heating a gas of uranium does not change the decay half life. Some think you can change this rate by feeding in KE (like X-rays) to change the decay half life of some isomers, but this is a contested viewpoint. and if applicable, it is adding the KE to the nucleus, not just speeding the total nucleus velocity in an external frame of reference. It is like the difference in firing a cannon ball. It speeds up, but the internal energy is not increased. For that you would need to heat the cannon ball separately.

Dan Tibbets

[Chris Bradley]

Dan DT wrote:
> Why does a mixture of Hydrogen and oxygen gas not spontaneously and rapidly burn to water. It is because a threshold has to be exceeded. This is often referred to as Gibbs free energy.

Dan, sorry but I think you need to revisit all of this. Re-hyopthesising interpretations you've made in your auto-didactic activities isn't helpful here, and this is word-salad. My thermodynamics is a bit shakey, but it looks like you are confusing activation energy with Gibbs free energy here. The rest I can barely even follow enough to comment on it - I cannot figure out what your point is.

[Steven Sesselmann]

Dan,

It is indeed fortunate that nuclei heavier and lighter than Ni62 don't spontaneously fall to the lowest potential state, well at least in the observers frame of reference.

You see it is all observer dependent, by definition the observer is always at ground potential (in my opinion around +930.412 MeV, the energy potential of Ni62). All the other elements have potentials either above us or below this, and will under the right circumstances attempt to fall towards ground potential. But thanks to the low permittivity of free space, these elements can't suddenly convert their potential energy to photons, so they hang around and wait for the right opportunity.

Now as you mentioned, it is believed that nuclear decay rates can not be influenced by external factors, but that's not really the case, we already know from relativity, that increasing the potential of matter, causes time dilation, which is in direct relation to it's decay rate.

Increasing the potential of matter is easy, you either accelerate it to a higher velocity, elevate it in a field, or move it to a large distance (into the past), all of which increases it's potential with respect to you the observer.

So by accelerating deuterons to large velocities in the hope that this will help to make them fuse, is to further increase the potential of a nucleus that is already suffering from a high potential, in other words pretty stupid!

Which then leaves us in a bit of a pickle, because to make deuterons easier to fuse we need to lower their potential, and trust me that is not so easy, how do you lower the potential of a deuteron already at rest?

This problem gave me quite a few sleepless nights, but there had to be a way, and one night the solution woke me up as if someone slapped me in the face...

For now you are going to have to figure it out for yourself, but soon I hope to demonstrate that man can control fusion, but I fear that fusion might end up controlling man, and that's pretty scary.

Steven

[Chris Bradley]

Steven Sesselmann wrote:
> Now as you mentioned, it is believed that nuclear decay rates can not be influenced by external factors, but that's not really the case, we already know from relativity, that increasing the potential of matter, causes time dilation, which has a direct relation to it's decay rate.
....
> So by accelerating deuterons to large velocities in the hope that this will help you make them fuse, is to further increase the potential of a nucleus that is already suffering from a high potential, in other words pretty stupid!

Sorry, Steven, that does not follow. The time dilation on a 20keV deuteron relative to a 1eV deuteron is negligible and does not account for why it is sooo much more likely to react than the 1 eV deuteron. You'll never get a fusion with such low energy deuterons, yet the time dilation only accouns for a few fractions of percent difference. You will have to come up with another explanation for justifying the 'high potential' argument, wrt DD fusion rates.

You'll also have to account for the fact that at higher energies still, the reactivity drops off. If 'higher potential' is gained by velocity, then how does this figure?

[Carl Willis]

This discussion is losing focus...heading off into La-la Land...

[Steven Sesselmann]

[Chris Bradley Wrote]
Sorry, Steven, that does not follow. The time dilation on a 20keV deuteron relative to a 1eV deuteron is negligible and does not account for why it is sooo much more likely to react than the 1 eV deuteron. You'll never get a fusion with such low energy deuterons, yet the time dilation only accouns for a few fractions of percent difference. You will have to come up with another explanation for justifying the 'high potential' argument, wrt DD fusion rates.

You'll also have to account for the fact that at higher energies still, the reactivity drops off. If 'higher potential' is gained by velocity, then how does this figure?[end quote]

Chris, my ramblings must have been a little confusing, but not quite La La land

I wasn't suggesting that time dilation had anything to do with the fusion rates, what I stated was simply that...

a) Lighter elements are naturally above ground potential
b) Observers are always at ground potential
c) Heavy elements are naturally below ground potential

From this it follows that lighter elements are more inclined to fuse when taken below ground, and heavier elements are more inclined to fission when taken above ground.

That (in my opinion) is why fusors work, and as you know I intend to prove it soon.

Steven

PS: Chris, from my logic above, it also follows that fusion reactivity should fall off with increasing particle energies, ie higher potential.

[ab0032]

Carl Willis:
> This discussion is losing focus...heading off into La-la Land...

Then let me try to bring it back to what I wanted to know.

What if we had an ideal 100% tranparent grid inside, that ions could not hit, what would happen? Would we still see fusion? Or would the plasma simply cool off? Is the grid required to absorb ions, so they can be replaced by new high energy ions?

Cheers,
Alex

[Frank Sanns]

Having no inner grid at all works great. The "Pillar of Fire" with plasma electrodes instead of an inner grid works very well and is patented. Us patent #7550741.

Frank Sanns

[Dan Tibbets]

Steven Sesselmann. Your assertions that nikel 62 is the zero point and fusion and fison exothermic reactions are either pos or negative ( greater or less potential energy) is not accurate. Nickel 62 has the smallest potential energy of any possible nucleus. A proton is not bound so has no potential energy defined . Actually the potential energy often is expresed as the sum of the repulsive energy (electromagnetic repulsion of protons), and the attractive energy of the Strong force. This attractive energy is by convention given a negative value (just like gravity). At Nickel62 the difference between these two forces is at a maximum( the most negative sum is reached). Past Nickel 62 the electromagnetic repulsion is growing faster than the Strong force attraction, as such the potential energy inside the nucleus is increasing (becoming less negative). At some point the net potential energy becomes maximum (~0) and the nucleus quickly falls apart. The position of Nickel 52 at the lowest potential energy state (most stable or most compact) is obvious from the Nuclear Binding Energy per Nucleon graphs. It is even more obvious if you flip the graph upside down (as it would be with a negative value assigned to the attractive Strong force component).

When light nuclei fuse you are harvesting the excess strong force energy. When you fission heavy elements you are harvesting the excess electromagnetic energy (compared to Nickel 62). The missing mass is important, but even more important is the balance of the two forces that make up this missing mass. primarily the repulsive Electromagnetic and attractive Strong forces.

I don't know how this would affect your relativity arguements . If we are at the baseline, increased speed of particles to relativistic speeds would not change their decay or reaction ratesin their frame of reference. From our frame of reference the reaction rates would seem to slow down - a good example is the survival of the Muon as it reaches the Earths surface. Without relativistic effects it would decay much too fast to survive it's several hundred mile trip through the atmosphere, from where it was born to the surface of the Earth. This would seem to be the reverse of what you are proposing. Traveling slower than us would result in faster reaction as percieved from our frame of reference, but how to do this ?

Dan Tibbets

[Chris Bradley]

Dan DT wrote:
> the potential energy often is expresed as the sum of the repulsive energy (electromagnetic repulsion of protons), and the attractive energy of the Strong force. This attractive energy is by convention given a negative value (just like gravity). At Nickel62 the difference between these two forces is at a maximum( the most negative sum is reached). Past Nickel 62 the electromagnetic repulsion is growing faster than the Strong force attraction

If the strong force does not balance the electromagnetic force perfectly, in any extant nucleus, then what other forces are at work to hold the nucleus together?

This just doesn't add up (whichever sign you attribute to each!!)

[Steven Sesselmann]

With all due respect to 300 years of science, if it can't answer Alex's simple question it's no b...dy good.
.

I would like to ask Alex a question, what is hypothetically the most negative charge a grid can have?

Steven

[Steven Sesselmann]

Dan Tibbets Wrote...
"Traveling slower than us would result in faster reaction as percieved from our frame of reference, but how to do this ?"

Dan..., you are knocking at the door, now you just have to find the key.

Steven

[Dan Tibbets]

Chris Bradly, i t seems few understand the Binding energy per nucleon relatio0nship (I cannot guarentee I'm not included in that group).

When the electromagnetic repulsion balances against the strong attraction, then the nucleus is balancing on the cusp of falling apart. This condition is approached as heavier nuclei are reached. It becomes prevalent beyond lead and just past plutonium there are no isotopes that can hold themselves together for long (with the strong force). Of course it varies with neutron count due to weak force and Paul exclusion processes, nucleon paring, etc, but the gross balance competition is between the strong force and the electromagnetic force (with a few variations like helium 4). The most stable, most dense nucleus, and the nucleus with the least potential energy is Nickel62. This is not where the two forces balance out. This is where the strong force has the greatest dominance. A nucleus that is ~ 4 nucleons wide is reaching the limiting range where the strong force will increase the attractive binding energy ( negative potential energy) by a greater amount than the acumulating repulsive electromagnetic repulsionby only a relative small amount compared to light nuclei where ALL of the nucleons are close enough together for the very short range strong force to reach across them all. With increasing nuclear diameter, the neighboring nucleons still attract each other, but not so much with nucleons on the other side of the nucleus. It is different for the electromagnetic repulsion because of the greater range of the force. I might add that using a negative value for the attractive force is a convention, but it is not necessary, so long as you keep track of the opposing effects. The attractive energy builds up , but saturates. The repulsive force continuously builds up and it saturates also, but less rapidly. A graph of the two opposing forces have different slopes and at Nickel 62 they cross. Beyond that point the repulsive force increases faster than the attractive force, but the attractive force had a head start because it was increasing faster (becoming more negative by convention) with the lighter nuclei. It takes it a while for the repulsive force to catch up (heaviest elements).
The Nickel 62 is the turn around point, the lowest potential energy possible, but that does not mean it is the point where a balance in established. It is the point where the trends reverse.
Another analogy. A balance scale. One one side is a small pan (representing the volume where the strong force can collect) and a much bigger pan on the opposite side (which represents the much larger volume where the repulsive electromagnetic force can collect).
The strong force lead pellets are added to the small pan at the same rate as lithium pellets are added to the large pan. Initially the scale will swing to the left side( strong force pan side) But the pan is small, additional lead pellets will fall over the side. Now, as additional pellets are added, the lithium pellets are still accumulating and the scale starts swinging to the right. At some point a balance will be reached where the arms are level. This would be ~ lead or beyond. The Nickel 62 is where the lead pan is filled up. This is a simplified analogy. It doesn't take into account that the strong force always continues to increase in a logarithmic fasion ( as does the electromagnetic force, but since this increase becomes progressively less with each step, it approaches a level line (like additional lead pellets falling out of the pan).

Dan Tibbets