"Science vs. Religion: What Scientists Really Think"

[polymath257]

humble brother wrote:

<quoted text>
So?
We're talking about a different model where all masses are different.

Several comments:
1) Since you model has dimensions of m^3/s^2 for M and m, it is better NOT to call them mass, but something else, say miasma.

2) Since F=ma, the dimensions of F are not in Newton's any longer, but are now m^4/s^4. So, again, it is better NOT to call it a force but something else, say a farce.

3) Your equation m=M^2/(v^2*r-M) is of no use unless there is an independent way to find M. So, how can we determine M for something like the sun? or a liter of water?

4) What happens when there are three gravitating bodies? four?

5) In your equation, m=M^2/(v^2*r-M), the v and r are orbial velocity and radius. Yoet you claim it shows how spin affects gravity. Could you supply details?

6) You claim this eliminates the need for dark matter and dark energy. Could you provide a detailed treatment of the rotation curve for some galaxy? How about a treatment of accelerated expansion?

[polymath257]

humble brother wrote:

The Earth is the arbitrarily chosen lump and within the new model the Sun is 1.68*10^5 times the size of that lump when talking about mass. We can call that earth-grams.
The Sun is 1.68*10^5 eg in mass, are you happier with that?

Please demonstrate how you arrive at this in your model.

[polymath257]

humble brother wrote:

<quoted text>
Oh, you're confused again. Chimney was calculating forces and you present numbers on acceleration.
It's all the same, eh?
Chimney's calculation, when corrected, reveals a considerable error.

The force depends on the mass of the object. For a 1 kg object, you get the same values for the force (in N) and the acceleration (in m/s^2) since F=ma.

For a 2 kg mass, the force would be doubled.

[polymath257]

humble brother wrote:

<quoted text>
You're going in circles. The original question was about how you can mathematically express the counter effect that rotation has to gravity.
Can you mathematically explain through your gravity models how the rotation of Earth counters some of the gravitational acceleration at the equator?
Have anything?

This is, yet again, an elementary physics problem appropriate for high school students. The force of gravity is F=GMm/r^2. The centripetal acceleration is v^2/r for a force of mv^2/r. For the equator, these are in opposite directions, so the total force F=GMm/r^2 -mv^2/r. Here, v is the rotational velocity (not an orbital velocity). For locations other than the equator, the forces are not in the same direction, so a vector subtraction has to be done. Again, an elementary problem for beginning physics students.

[Kong_]

Poly....saw this and thought of you:

http://asset.soup.io/asset/0305/8923_4237_500...

Enjoy!

[polymath257]

humble brother wrote:

The Earth is the arbitrarily chosen lump and within the new model the Sun is 1.68*10^5 times the size of that lump when talking about mass. We can call that earth-grams.
The Sun is 1.68*10^5 eg in mass, are you happier with that?

What is that in units of m^3/s^2?

[Chimney1]

humble brother wrote:

You are quite confused. Apparently the size of the arbitrarily chosen relative lump for a common measure has some emotional value for you.
Instead of the kilogram lump for this I prefer to use the whole Earth as the lump. Do you understand that?

Yes, but do you understand why that won't work? I will paint it out for you (again). The arbitrary lump has to be of a magnitude we can measure other properties for, directly, to learn more about it. For example, we can measure the force required to accelerate one lump of a kg by 1 meter per second per second (and then call that force 1 Newton). We can directly measure the gravity force generated by one large ball of say 1000kg to another (as Cavendish did). But we cannot accelerate the earth and sun at will, or directly measure the forces between them, which is why its a useless base measure of inertial mass. Its a practical question. And, using earth as a base measure, we could not even determine the relative mass of that 1kg to it...

Let me put it in practical terms for you. If I decided to accelerate a kg of material out of the solar system, we can calculate the actual fuel requirements and rocket power needed based on the actual force in our baseline units etc. Knowing the energy that our fuel an generate, etc, in compatible units, we can know in absolute terms how much fuel is required, the power generation of the rocket, thus the trajectory, etc.

However, we could not do that using the earth as a baseline. We do not know the absolute quantity of force required to move the earth at a particular acceleration, and relate it to actual measurements of the energy and power generation of fuels and rockets. Thus with "1 earth" as a unit of mass, we are stuck with mere relative masses that we can never tie in with other units or derived units.

Since you do not understand the difference, I expect you will not understand my point. So just go ahead and tell me how, using your "1 earth" as a mass, we can measure the relative mass of that lump of platinum iridium to it. Using our method, and G, we can. Using yours, its not possible.

[Chimney1]

humble brother wrote:

<quoted text>
Oh, you're confused again. Chimney was calculating forces and you present numbers on acceleration.
It's all the same, eh?
Chimney's calculation, when corrected, reveals a considerable error.

Wrong. My numbers used the approximation of 9.8 N and 6380km radius and 440 m/s/s for the speed at the equator. You will see that AM's number show a similar difference between the baseline and the equatorial centripetal effect. If I was going to do the calculation with high precision, I would use more significant places for the inputs AND use a slightly different radius for the poles compared to the equator.

However, I was merely demonstrating the equations and calculations at work, and as an estimate, we can happily use the difference of 0.03 newtons as a ballpark figure, demonstrating the principle.

And AM understands, as you appear not to, the direct connection between force and acceleration on a given mass. I am sure you know that F = ma, but you do not really know what that means in physics terms. Remember, we even had to teach you that mass is not weight, and you STILL do not get the difference between relative and absolute mass.

Now, show how your equations can derive a similar figure and also how you can determine the mass of the earth and sun, and relate that mass to the lump of platinum iridium in Paris or for that matter the mass of an elephant.

You cannot, you never could, and a month of your ridiculous evasion and lies has not changed that.

[Chimney1]

polymath257 wrote:

<quoted text>
What is that in units of m^3/s^2?

All HB's computations of the farcical miasma lead to a singularity of theoretical absolute density.

[humble brother]

polymath257 wrote:

Fine, so talk about the wavefront rather than a specific photon.*sigh*

In the double slit experiment when photons are shot "one at a time" they register at the sensitive camera's CMOS as a single pixel events, not as blurred blobs of multiple pixels. Every pixel in the camera's CMOS can be considered as a separate observer, and only one of them will catch a single photon.

How could a wave front end up only in one pixel on the CMOS?

polymath257 wrote:

Yes, the speed of light is constant in all reference frames and has the same value in all of them. Nothing you have said contradicts this.

Haha.. How about this:

polymath257 wrote:

The light has moved 5 light seconds and the ships have both moved 2.5 light seconds in that 5 seconds. So the light has already passed one and has yet to pass the other.

There is no relative to light movement for the observers, only light moves towards them. Do you not understand that?

Both ships are 5.0 seconds (emitter time) away from the beacon when first light gets emitted.
Both ships are 4.35 seconds (ship time) away from the beacon when first light gets emitted.

You are constantly giving that light a different speed for different ship!
The ships do not move relative to that light. Once it is emitted it moves at the same observed speed for both ships!

[humble brother]

polymath257 wrote:

Several comments:
1) Since you model has dimensions of m^3/s^2 for M and m, it is better NOT to call them mass, but something else, say miasma.

2) Since F=ma, the dimensions of F are not in Newton's any longer, but are now m^4/s^4. So, again, it is better NOT to call it a force but something else, say a farce.

3) Your equation m=M^2/(v^2*r-M) is of no use unless there is an independent way to find M. So, how can we determine M for something like the sun? or a liter of water?

4) What happens when there are three gravitating bodies? four?

5) In your equation, m=M^2/(v^2*r-M), the v and r are orbial velocity and radius. Yoet you claim it shows how spin affects gravity. Could you supply details?

6) You claim this eliminates the need for dark matter and dark energy. Could you provide a detailed treatment of the rotation curve for some galaxy? How about a treatment of accelerated expansion?

1) I was actually thinking about this earlier. Mass in the old concepts seems to indicate the amount of substance and not the amount of gravitational effect (relative to acceleration).

I considered whether this new concept should be called "nass" :)
I could call it that. If you double the nass (mass effect) of an object it significantly affects the gravitational acceleration. If you double the substance (mass) of the smaller object mass effect (nass) does not double.

2) You have your units stuffed in your magical variable G so that things appear to work. Would you like to explain HOW your units actually ended up in G?

3) You make Earth (m) the arbitrarily chosen lump used as a relative measure. Then you express the mass of the Sun relative to that chosen measure.

4) You have to ask? They all pull each other, the calculation of forces is just a bit more complex.

5) You can see from the equation that mass is subtracted from v²r so they have the same unit. Any spinning mass can be thought of just multiple joined masses.

6) You expect me to do all your work for you?:D
If you want to test it, then test it. You have the equations.

This new model actually forces me to change my earlier hypothesis on the accelerated expansion of the universe. This model indicates that as stars steadily lose their mass their gravitational pull decreases and thus everything is moving apart. In the scope of the solar system the planets also capture some of the mass that the Sun loses.

[humble brother]

Chimney1 wrote:

<quoted text>
Yes, but do you understand why that won't work? I will paint it out for you (again). The arbitrary lump has to be of a magnitude we can measure other properties for, directly, to learn more about it. For example, we can measure the force required to accelerate one lump of a kg by 1 meter per second per second (and then call that force 1 Newton). We can directly measure the gravity force generated by one large ball of say 1000kg to another (as Cavendish did). But we cannot accelerate the earth and sun at will, or directly measure the forces between them, which is why its a useless base measure of inertial mass. Its a practical question. And, using earth as a base measure, we could not even determine the relative mass of that 1kg to it...
Let me put it in practical terms for you. If I decided to accelerate a kg of material out of the solar system, we can calculate the actual fuel requirements and rocket power needed based on the actual force in our baseline units etc. Knowing the energy that our fuel an generate, etc, in compatible units, we can know in absolute terms how much fuel is required, the power generation of the rocket, thus the trajectory, etc.
However, we could not do that using the earth as a baseline. We do not know the absolute quantity of force required to move the earth at a particular acceleration, and relate it to actual measurements of the energy and power generation of fuels and rockets. Thus with "1 earth" as a unit of mass, we are stuck with mere relative masses that we can never tie in with other units or derived units.
Since you do not understand the difference, I expect you will not understand my point. So just go ahead and tell me how, using your "1 earth" as a mass, we can measure the relative mass of that lump of platinum iridium to it. Using our method, and G, we can. Using yours, its not possible.

So can you actually explain the logic of the units ending up in G?
Why are those exact units stuffed in G?

[humble brother]

Chimney1 wrote:

All HB's computations of the farcical miasma lead to a singularity of theoretical absolute density.

Lets look at your model..

F=ma clearly shows that if you double the force affecting a mass you DOUBLE the acceleration.
F=GMm/r² does NOT even begin to satisfy that.

Above we see that you can double the force by doubling M or m.
You could double the gravitational acceleration on Earth by doubling the mass of Earth.

Apparently gravitational force for a 2 kg lump is twice that of a 1 kg lump.
The gravitational force is doubled, why is not the acceleration doubled?

In reality your model is a miserable failure of gibberish in the language of mathematics.

[humble brother]

havent forgotten wrote:

that was to am, and this is off topic. just to say hi and check in on our mutual acquaintance! has he apologized yet? are you really in Finland, by the way? and what nationality are you? As I watch Olympics I pay more attention to nationality of folks on topix, but some could be Amricans abroad.

The person behind the alias Skeptic has not uttered much here. Perhaps a couple of times he has had to shout out his opinions when the apparent urge to do that has won.

Yes, I'm in Finland and I am also Finnish. My wife is originally Canadian though :)

[Chimney1]

humble brother wrote:

<quoted text>
Lets look at your model..
F=ma clearly shows that if you double the force affecting a mass you DOUBLE the acceleration.

Correct.

F=GMm/r² does NOT even begin to satisfy that.
Above we see that you can double the force by doubling M or m.

Yes - and there is nothing wrong with that...

You could double the gravitational acceleration on Earth by doubling the mass of Earth.

No, you would double the gravitational FORCE, not the gravitational ACCELERATION. Two different things...

Apparently gravitational force for a 2 kg lump is twice that of a 1 kg lump.

Correct!

The gravitational force is doubled, why is not the acceleration doubled?

Because when you double the mass, you double the inertia (resistance to acceleration). So twice the force acting on twice the mass = the same acceleration.

In reality your model is a miserable failure of gibberish in the language of mathematics.

As the above extremely elementary example shows, its you who needs to understand the basics. You appear to be making the mistake that gravity is a uniform force acting on all objects, i.e. the same force acting on 2kg as on 1kg. But that is not the case. Gravity acts uniformly PER UNIT OF MASS. There IS twice as much total force on a 2 kg object as a 1kg object in the same gravity field, but the same amount of force per unit of mass.

[Chimney1]

humble brother wrote:

<quoted text>
So can you actually explain the logic of the units ending up in G?
Why are those exact units stuffed in G?

Yes. But you will follow only if you read carefully and do not go off half cocked by the second line, as you are prone to do.

Force is a unit derived the fundamental unit MASS and the derived unit ACCELERATION which itself is derived from the fundamental units of DISTANCE and TIME, as follows:

F = ma

F = mass * distance / time / time.
i.e. F = m * d * t^-2

When deriving an expression for the FORCE of a gravitational attraction for two objects, the contributing factors must have a unit expression that is the same as force. BTW, "m" is for mass, not meters, so don't mix them up a second time.

Newton recognised that the force varied as a function of the mass of each object and the distance between the objects squared.(Twice the distance, a quarter the attraction), by a factor of some discoverable constant representing the actual field strength of a G field which was not derivable from pure math but was a discoverable property of the universe. Requiring empirical observation to pin down. However, the units of G were easily derived from the pure equations:

If gravity is an effect of M1 * M2 / r or d (radius, or distance between centres of mass)^2. The units are:

F = m * d * t^-2 but =/= m1 * m2 * d^2

To derive FORCE from the right hand expression, you need to find the ACTUAL expression of field strength applying to gravity in our universe. That is G!!! And G must be expressed in the units that balance the equation as follows:

F = m * d * t^-2 = m1 * m2 * d^-2 * "d^3 * m^-1 * t^-2" (which is G)

i.e. the expression for the actual force generated by a given amount of matter (mass) at a given distance must be expressed in the units to the far right of the equation that I put into "", multiplied by m1 * m2 * d^-2, to give a resultant FORCE that must be expressed as m * d * t^-2

That is the explanation, given from two sides. On one hand, the equation's units must balance. But that is secondary. The primary point is, in physics terms, that to quantify the total force in a given gravity system, you have to know the distance and masses AND know the actual empirically discoverable field strength that any given masses will generate over any given distance i.e. G. Its a real property of our universe (and still valid with Einstein too).

[humble brother]

Chimney1 wrote:

Correct.
<quoted text>
Yes - and there is nothing wrong with that...
<quoted text>
No, you would double the gravitational FORCE, not the gravitational ACCELERATION. Two different things...
<quoted text>
Correct!
<quoted text>
Because when you double the mass, you double the inertia (resistance to acceleration). So twice the force acting on twice the mass = the same acceleration.
<quoted text>
As the above extremely elementary example shows, its you who needs to understand the basics. You appear to be making the mistake that gravity is a uniform force acting on all objects, i.e. the same force acting on 2kg as on 1kg. But that is not the case. Gravity acts uniformly PER UNIT OF MASS. There IS twice as much total force on a 2 kg object as a 1kg object in the same gravity field, but the same amount of force per unit of mass.

Your whole logic falls apart in this simple scenario:
Suppose a massive spaceship travels to Earth. The mass of this spaceship is Exactly the same as the mass of Earth.

The ship does not go into orbit but moves directly towards Earth and uses its engines to decelerate and enters the atmosphere. When the ship reaches the altitude of 1.0 kilometers it powers its engines so that the movement towards Earth stops. So by using energy the ship hovers in the air.

Suddenly the engine of the ship dies and it begins a free fall.

What is the relative acceleration that an observer standing on Earth will measure for that falling ship that has the same mass as Earth?

[humble brother]

What is the acceleration if that ship's mass is:

-50 % of Earth's mass
-25 % of Earth's mass
-10 % of Earth's mass
-5 % of Earth's mass
-1 % of Earth's mass

???

[humble brother]

Chimney1 wrote:

BTW, "m" is for mass, not meters, so don't mix them up a second time.

That's funny. You are the one who confused units with variables of the equations.
Would you like me to quote your post and provide the link to that failure of yours?

[humble brother]

Chimney1 wrote:

To derive FORCE from the right hand expression, you need to find the ACTUAL expression of field strength applying to gravity in our universe. That is G!!! And G must be expressed in the units that balance the equation as follows:
F = m * d * t^-2 = m1 * m2 * d^-2 * "d^3 * m^-1 * t^-2" (which is G)

"To balance the equation" :D

1. you have a falsifiable equation
2. you need to balance it or it is a fallacy
3. you add units to make the equation balanced

So you have a falsifiable equation which you needed to balance. Do you understand that the balance is as falsifiable as the model?

Chimney1 wrote:

i.e. the expression for the actual force generated by a given amount of matter (mass) at a given distance must be expressed in the units to the far right of the equation that I put into "", multiplied by m1 * m2 * d^-2, to give a resultant FORCE that must be expressed as m * d * t^-2
That is the explanation, given from two sides. On one hand, the equation's units must balance. But that is secondary. The primary point is, in physics terms, that to quantify the total force in a given gravity system, you have to know the distance and masses AND know the actual empirically discoverable field strength that any given masses will generate over any given distance i.e. G. Its a real property of our universe (and still valid with Einstein too).

As you have proven now G is not a property of the universe but falsifiable with your model.
G is your very precious magic constant.