[Bored chemist]

Are you sure that this formula is correct?

Yes we are.
And the reason why is it correct is given in the article cited.
It's the derivative of the gravitational force wrt distance.


StatementOfObviousForDave.JPG (35.46 kB . 711x214 - viewed 941 times)

[Dave Lev (OP)]

You have got that wrong too.
With a  single ended amp the best power I can get to the load has a peak to peak voltage equal to value of the supply voltage.
With a push-pull amp, I can get twice the peak voltage.
So the power available is four times greater.
1+1 is not 4.

We do not get those push pull waves in the same time.
Please remember, we discuss about heat/power/energies and not about voltage.
In the same token, it is not about pure tidal gravity force, but about tidal heat dispassion.
Don't you agree that in the push pull amplifier we should add the energy of the positive wave to the energy of the negative wave.
Hence, the Sum of the energies of the push and pull waves indicates about the total energy of the wave.

The Tidal power/energy is working exactly in the same way as the Push pull amplifier.
However, while the amplifier Push activity works at different times from the Pull activity, the Tidal push pull activities work simultaneously.
Therefore, while the Pull tidal gravity force increases the tidal heat/energy/power of the left side of the object, the Push tidal gravity force increases tidal heat/energy/power of the right side of the object.
Hence, we actually double the heat energy in the object at the same time.
It is so simple and clear.
So, why do you insist to deduct the tidal heat energy from one side with the tidal heat energy of the other side?

So, why the science community have decided to ignore all the internal tidal forces, including the tidal push /pull mechanism, and used the following formula:

Because tides are not the same as gravity.
Why do you not realise that?
You have been told several times.

Again, we must focus on the total tidal heat energy and not the reference between the tidal forces.
In any case, tidal force is based on gravity force
Take out the gravity from the tidal and you have no tidal.

Quote from: Dave Lev on Today at 07:00:28
Are you sure that this formula is correct?

Yes we are.
And the reason why is it correct is given in the article cited.
It's the derivative of the gravitational force wrt distance.

We can't just deduct the tidal gravity forces and then extract the total tidal energy.
We must first find the tidal heat energy due to the tidal gravity forces on each atom and then add them all together.
Again, we focus on the total tidal heat dissipation/power and energy in the object
Therefore, it is a severe mistake to deduct tidal forces before we discover the impact of the tidal heat on every atom in the object.
Let's use the following example about pot with boiling water.
On the left side of the pot, we would set red water.
On the right side we would set blue water. (let assume that they do not mix together)
Therefore, do you really think that the total heat energy in the Pot is the deduction of the heat energy between the red water to the Blue water?
Therefore, how can we claim that the total heat dispassion due to tidal gravity force is the deduction of the tidal heat dissipation from one side of the object with the tidal heat dispassion from the other side of the object.

[Bored chemist]

So, why do you insist to deduct the tidal heat energy from one side with the tidal heat energy of the other side?

We  don't.

If a force to 1000 Newtons acts on an object from one side and a force of 999 Newtons acts on the opposite side and the object moves 1 metre how much work is done on the object?
Do you understand that the answer is 1 Joule, or do you hallucinate that it's 1999 Joules?

Please remember, we discuss about heat/power/energies and not about voltage.

Push Pull drive lets you double the voltage swing, so it gets you four times the power.
But your being wrong about doesn't matter much because it is irrelevant.

In any case, tidal force is based on gravity force

No. It is based on the first derivative of gravity wrt distance.

[paul cotter]

Ignoring saturation voltages a single ended amplifier and a push -pull amplifier will both output a peak to peak voltage equal to the supply rail. A bridge push pull will output twice the supply rail voltage and four times the power output, all other parameters unchanged. All above in the absence of transformers. The basic push pull is way more efficient than the single ended amplifier as the latter suffers from the lack of an active pull up/pull down mechanism.

[Kryptid]

Therefore, the total tidal force must be the sum of all the tidal forces on each atom in the object.

The equation takes that into consideration. The tidal force is measured in newtons per meter of distance. That would include all of the atoms in that same distance. As an example, let's say the tidal force is 10 newtons per meter and there are 1 trillion atoms in that meter. That would be an average force of 10^-11 newtons per atom. If you really wanted to, you could calculate the total average force on all of the atoms in a body rather easily.

[Dave Lev (OP)]

Thanks Kryptid
Do appreciate this clear explanation.
Now I do understand what is the source of the problem.
The tidal force formula is perfectly OK.
I have even proved it to myself by using gravity force VS centrifugal force by just few steps. (If you wish I can introduce this calculation.)
"The tidal force is measured in newtons per meter of distance."
The tidal force doesn't just lift the bulge it lifts every atom in the sun.

Let's go back to your calculation:

So now we will calculate the tidal force between the Sun and Alpha Centauri A:

(2GM1M2)/r3
((2)(6.674 x 10-11)(1.9885 x 1030)(2.1452 x 1030))/(4.1097 x 16)^3
((5.694 x 10^50)/(6.941 x 10^49)
8.2 newtons per meter

The total tidal force can be computed by multiplying the force per unit distance by the diameter of the Sun:

8.2 x 6.96 x 108= 5.7 x 10^9 newtons

The Diameter of the Sun is 1,392,000 km = 139,200,000 m
The radius is 69,600,000 m
Its volume in m is
V = 4/3 π R^3 = 4.18 * 337 10^21 m^3 = 4.18 * 337 10^21 m^3 = 11.5 10^24 m^3.
Even if we keep the tidal force between the Sun to Alpha Centauri A at 8.2 N

The total force is:
F = 8.2 * 11.5 10^24 = 9.43 10^25 N

Do you still claim that this tidal force is neglected?

[Bored chemist]

Even if we keep the tidal force between the Sun to Alpha Centauri A at 8.2 N

The total force is:
F = 8.2 * 11.5 10^24 = 9.43 10^25 N

That makes no sense.
The units are wrong.

[Origin]

Even if we keep the tidal force between the Sun to Alpha Centauri A at 8.2 N

This is incorrect, the tidal force is 8.2 N/m, not 8.2N.

The total force is:
F = 8.2 * 11.5 10^24 = 9.43 10^25 N

That is not correct.  To get the total force you multiply by the diameter, not the volume.
If you multiply this by the diameter of the sun you will get the total force: 1.14 x 10^10N.
Your incorrect calculations gives:   9.43 10^25 N-m^2, which is not force.
You cannot arbitrarily change an equation in physics and expect to get a reasonable answer

Do you still claim that this tidal force is neglected?

I think we all agree the force is negligible.

[Janus]

That is perfectly OK
The oort cloud extends up to 3.2LY away from the Sun and sets the border of the solar system.
Therefore, it is very clear that even if we just focus on the 400 stars that are located up to 20 LY away from the sun (and ignore all the other millions that are located further away) we should discover that they set severe tidal heat forces on the Sun while each one works from different location.

Except that's not how any of this works.  Masses spread out surrounding the Solar system would cancel each other's tidal influences on the Sun out, they would not add up.  We see this same thing when the Sun and Moon are at right angles to each other in the sky. We get less intense tides as they work against each other.
In addition, tidal forces increase across the size of the region they are acting.  Therefore, tidal forces powerful enough to produce the type of heat energy the Sun outputs would rip the planets away from the Sun. The Sun's Hill sphere would be too small to allow for it to hold on to orbiting bodies

[Kryptid]

The total force is:
F = 8.2 * 11.5 10^24 = 9.43 10^25 N

Do you still claim that this tidal force is neglected?

Newtons per meter is a measure of force along a unit of length, not a unit of volume. So your math is incorrect.

[Dave Lev (OP)]

Newtons per meter is a measure of force along a unit of length, not a unit of volume. So your math is incorrect.

Dear Kryptid
Please look again in the following diagram:

Please see the following diagram:
https://www.wolframcloud.com/objects/demonstrations/TidalForces-source.nb
We see that the tidal force works on every atom in the Object.
In the left side of the object, every atom is pulled to the left, while in the right side of the object every atom is pushed to the right.

Don't you agree that we clearly see that the tidal force works on every atom in the Object.
You have even stated that the tidal force "include all of the atoms in that same distance":

The tidal force is measured in newtons per meter of distance. That would include all of the atoms in that same distance. As an example, let's say the tidal force is 10 newtons per meter and there are 1 trillion atoms in that meter. That would be an average force of 10-11 newtons per atom. If you really wanted to, you could calculate the total average force on all of the atoms in a body rather easily.

So how can you claim now that only the atoms in the diameter are effected by the tidal force?
What about the atoms above & below the diameter?
Why do you insist that tidal force can't work on them?
If that was the case, then it is expected that only those atoms in the diameter would set the bulge.
Hence, as Io bulge height is 100 m, don't you agree that based on this message, it is expected to see a tower shape of 100m (that is ejected directly from the diameter)?
Is that what we see at Io?
Would you kindly reconsider your reply?

Why are all in panic to discover that tidal force works on every atom in the object?
What could happen to the sun if suddenly the science community would understand that the tidal force is good enough to generate its internal electromagnetic force?

[Bored chemist]

Please look again in the following diagram:

Looking at any diagram will not fix the problem with your meaningless calculation, will it?
The units will still be wrong, not matter who looks at what.

Why did you think looking at a diagram (one with which we are already familiar) would help somehow?

So how can you claim now that only the atoms in the diameter are effected by the tidal force?

Nobody did.
That's just your lack of understanding again.
It's not a matter of science, but of reading comprehension.

Why do you insist that tidal force can't work on them?

Again.
Nobody did.

Why are all in panic to discover that tidal force works on every atom in the object?

We are not.
It's just that you can't, or refuse to, understand what we say.

[Bored chemist]

What could happen to the sun if suddenly the science community would understand that the tidal force is good enough to generate its internal electromagnetic force?

You already know the answer to that.

Therefore, tidal forces powerful enough to produce the type of heat energy the Sun outputs would rip the planets away from the Sun. The Sun's Hill sphere would be too small to allow for it to hold on to orbiting bodies

[Kryptid]

So how can you claim now that only the atoms in the diameter are effected by the tidal force?

I didn't. All of the atoms are affected because all of the atoms are already within the diameter of the Sun.

Why are all in panic to discover that tidal force works on every atom in the object?

We're not. We already knew that it did.

What could happen to the sun if suddenly the science community would understand that the tidal force is good enough to generate its internal electromagnetic force?

But it isn't. Origin already pointed out that your equation results in the wrong units for the calculation to be measuring force. Let me elaborate on that. Let's go back to the 8.2 newtons per meter thing. That in itself is not a measure of force, but rather the measure of a force gradient (how force changes over distance). This means that every meter further from the gravitating body you get, the gravitational force acting on it becomes 8.2 newtons weaker. So that in itself doesn't tell you the total force acting on the body being affected by the gravity.

When you do math with measurements, you also have to do math with the units themselves. So if we want to know the total tidal force affecting the Sun, we need to somehow go from units of "newtons per meter" to regular old "newtons". In order to do that, we need to get rid of the meter part of the measurement (i.e. multiply 8.2 newtons per meter by the total number of meters over which the force gradient is being measured). This is because dividing a force by a length and then multiplying by a length again will go back to force, as multiplication and division are opposite functions and thus cancel out when performed in succession. So multiplying 8.2 newtons per meter by the diameter of the Sun (1,392,000,000 meters) yields a force of 1.14144 x 10¹⁰ newtons. That represents the difference in force between the near side of the Sun and the far side of the Sun from Alpha Centauri A.

When we do things your way, we end up multiplying a force gradient (newtons per meter) by a volume (cubic meters). This results, as Origin said, in newton-square meters. I don't know what that would even be measuring, but it's not a force. It's almost like confusing a pressure for a force, but it's also not a pressure. If you want your calculations to be valid, you need to get rid of the square meters and be left with nothing but newtons.

[Halc]

All of this is pretty pointless. Everybody's discussing tidal force, but tidal energy is very different from tidal force. For instance, Jupiter exerts incredible tidal force on its four larger moons, but those moons get almost zero energy from that The heating of IO comes from the other moons, far less tidal force, but far more tidal energy since IO isn't locked with them.

Tidal force is neither a scalar nor a vector. It's something like a vector with orientation but no direction. To be honest, the way to compute tidal force is simply to compute the gravitational field (potential) about the sun, taking everything into consideration, and then computing the 2nd derivative of that. Doing so gets you something entirely irrelevant to computing the energy the sun gets from those forces. This topic is supposed to be about energy, not force. Energy from tidal action on the sun is a function of its elasticity, rotation rate relative to the fields, etc

The energy output of the sun would have to be at least as large as the rate at which its angular kinetic energy is decreasing. Both of those are pretty easy to compute, far easier than trying to directly figure out how much tidal energy the sun gets from various objects around it. You'll find that the energy falls probably at least 20 orders of magnitude short of where it needs to be. Case close.

Dave, the math I suggest is trivial. Do it and show me otherwise.

[Bored chemist]

All of this is pretty pointless. Everybody's discussing tidal force, but tidal energy is very different from tidal force. For instance, Jupiter exerts incredible tidal force on its four larger moons, but those moons get almost zero energy from that The heating of IO comes from the other moons, far less tidal force, but far more tidal energy since IO isn't locked with them.

Tidal force is neither a scalar nor a vector. It's something like a vector with orientation but no direction. To be honest, the way to compute tidal force is simply to compute the gravitational field (potential) about the sun, taking everything into consideration, and then computing the 2nd derivative of that. Doing so gets you something entirely irrelevant to computing the energy the sun gets from those forces. This topic is supposed to be about energy, not force. Energy from tidal action on the sun is a function of its elasticity, rotation rate relative to the fields, etc

The energy output of the sun would have to be at least as large as the rate at which its angular kinetic energy is decreasing. Both of those are pretty easy to compute, far easier than trying to directly figure out how much tidal energy the sun gets from various objects around it. You'll find that the energy falls probably at least 20 orders of magnitude short of where it needs to be. Case close.

Dave, the math I suggest is trivial. Do it and show me otherwise.

Sorry Halc, but, if Dave was interested in getting the right answer, he'd have realised roughly what that maths would say when I pointed this out.

{Dave says} the tidal forces of distant objects provide (significant) heat in the sun.


But, to provide heat they have to do work.
Work is done when a force moves through a distance.
In this case, the force would be accelerating the part of the sun subject to tides and the distance would be the difference between "high tide" and "low tide" on the sun.
And as was pointed out, that would require the sun to shrink and grow in response to the tide effects.
But the diameter of the sun is essentially constant.

So we know it is not being stretched and squashed by tidal effects.
So we know that practically no work is being done on it.
So we know that work can't be the source of the sun's heat.

And, since he's not interested in rational debate or learning science, I think he's not going the site any good so we should ban him.

[Dave Lev (OP)]

Dear Halc

Everybody's discussing tidal force, but tidal energy is very different from tidal force. For instance, Jupiter exerts incredible tidal force on its four larger moons, but those moons get almost zero energy from that The heating of IO comes from the other moons, far less tidal force, but far more tidal energy since IO isn't locked with them.

Tidal force is neither a scalar nor a vector. It's something like a vector with orientation but no direction.  This topic is supposed to be about energy, not force. Energy from tidal action on the sun is a function of its elasticity, rotation rate relative to the fields, etc

I fully agree with your above explanation.
Thank you for refocus the discussion.
Let's look again on Io Moon.
We have measured a Bulge of 100 m.
Therefore, I hope that we all agree that by everage we can claim that every atom there had been lifted by 50M.
Hence the work that is needed to set that bulge is:
W = MGH.
Io mass = M = 8.93 10^22 Kg
G = 9.8 m/s^2
H = 50 m
W = 4.375 10^25
However, that work is useless if the moon is face locked with its planet (as our moon) where its spin angular velocity is fully correlated with its orbital angular velocity around the earth.
In this case, the Bulge is fixed, it doesn't move across the moon and it doesn't contribute any sort of heat.
Therefore, technically even if the tidal force was stronger by 1 M times, it won't contribute any energy as long as it is face locked with the planet.
Hence, we can't claim that tidal force is automatically translated into heat energy.
In order to get that heat, the Bulge must move around the moon mass and set a friction between its atoms. That friction is the real source for what is called "tidal heat".
However, there is a small problem.
If there is a friction in the atoms, then it should slow down the spin velocity of the Moon.
As this spin velocity decrease, then eventually it would be face locked with its planet.
So, we can ask, how could it be that most of the orbital objects are not faced locked with their host?
The science community assume that over time every orbital object must decrease its spin velocity.
I claim that this assumption is incorrect and somehow the tidal force is also responsible for the continuation of the orbital spin/rotation
So how it works:
Please look again at the following image:
https://www.wolframcloud.com/objects/demonstrations/TidalForces-source.nb
I would like to focus on the two main tidal forces.
One is horizontal tidal force and the other is a vertical tidal force.
The science community mainly focus on the horizontal tidal force as the source for the Bulge.
However, the vertical tidal force is neglected.
We actually do not even think about it.
I claim that this vertical tidal force is the MOST important force!!!
Without it, there will be no Dynamo and no magnetic fields.
So why does the vertical tidal force spin the Dynamo?
We can use a Spiral Ratchet Screwdrivers as nice example:
https://toolsforworkingwood.com/store/item/EE-SRS.XX
In this case, the vertical force is transformed into spin velocity.
However, it is clear that this example can't give the real physics answer.
Never the less, the Magnus effect might be the perfect explanation:
https://en.wikipedia.org/wiki/Magnus_effect
"The Magnus effect is an observable phenomenon commonly associated with a spinning object moving through a fluid. A lift force acts on the spinning object."
In this case, a spinning object moving through a fluid would generate a lifting force.
However, in our case, we discuss about an object (dynamo) that moves in a fluid while the vertical tidal force works against its natural lifting force.
Therefore, there is a possibility that as we reverse the conditions and offer the force that works against the natural lifting force, this object (dynamo) would spin.
Conclusion:
The tidal horizontal force works on the spinning object and increase its internal heat due to internal frictions. At high enough temp the internal atoms would be at fluid mode.
Never the less, due to the mighty gravity pressure, the internal core would be transformed into a solid core.
However, in order for that solid core / dynamo to work, it must spin much faster than the self' object spin.
Therefore, somehow an external force is needed for that spinning motion.
That force is called the vertical tidal spinning force.
Without it, the dynamo won't spin faster than the object' spin and it won't generate any sort of magnetic field.
So, in reality, the heat due to the tidal force isn't translated into any EM energy.
It is just used to set the internal matter in a fluid mode in order to give the dynamo the possibility to spin, while the vertical tidal force is the one that force the dynamo to spin at higher velocity (with reference to the object spin).
Therefore, this vertical tidal force is the only force in the nature that can force the dynamo to spin at high velocity.
Take out the vertical tidal spin force and the Dynamo would just spin at the same velocity as the object itself without generating any sort of magnetic fields.
The Fusion activity idea can increase the internal heat but it can't force the dynamo to spin at higher velocity than the object itself.
Therefore, without vertical tidal spin force there is no way to spin the dynamo and gain any sort of magnetic fields.

[Origin]

Seriously Dave?  Are you really going to double down on this silliness?  You must be trolling.
I agree with bored chemist, he should be banned as a troll.  I don't think he is as stupid as he acts,  he is just trying to annoy members.

[Kryptid]

In this case, the Bulge is fixed, it doesn't move across the moon and it doesn't contribute any sort of heat.
Therefore, technically even if the tidal force was stronger by 1 M times, it won't contribute any energy as long as it is face locked with the planet.
Hence, we can't claim that tidal force is automatically translated into heat energy.

There are two things you are missing. One, Io has an eccentric orbit. That means that it is closer to Jupiter at some times and further away at other times. This therefore changes the size of the tidal bulge over time. This up-and-down flexing of the Moon generates heat. Two, as Halc stated, Io also receives tidal heating by interacting with Jupiter's other moons.

I claim that this vertical tidal force is the MOST important force!!!
Without it, there will be no Dynamo and no magnetic fields.

Provide a citation for this claim, please. If that was true, then we would be incapable of building artificial dynamos on Earth because we can't create tidal forces at will. The fact that we can proves that you are wrong.

In this case, a spinning object moving through a fluid would generate a lifting force.

You misunderstand the Magnus effect. In order for it to work, the fluid moving across one half of the spinning object has to be moving faster than on the other side. That isn't what happens in the cores of planets.

However, in order for that solid core / dynamo to work, it must spin much faster than the self' object spin.

Give a citation for this claim as well. I'm going to start bearing down on you to give us these citations. I'm really starting to consider locking this thread if you keep making claim after claim without actually backing them up with a reputable source. If you want to avoid that, then get to citing your sources.

You still haven't explained the neutrino problem nor given any correct math to show that your idea can work.

[Bored chemist]

G = 9.8 m/s^2

When you say things like that, is it because you don't know that it makes you look stupid, or that you don't care about how stupid you look?