I’ve made a series of the biggest discoveries in physics.

[DrP]

1 minute ago, OlegGorokhov said:

The 2nd second of the movement of the stone (under the action of the pushing Carlson) and the 2nd second of free fall of any stone (under the action of the pushing gravitational force) are completely identical processes.

These two processes are completely identical, because 

I don't think they are identical because KE = 1/2 MV^2    -  so the faster you go the more energy required to accelerate. In the case of gravity the force increases as you get closer to the ground as force is directly proportional to the inverse of the separation distance SQUARED.

You must know this though - what am missing now?

 

[OlegGorokhov]

20 minutes ago, DrP said:

The force from gravity increases the closer the objects get via an inverse square law. The KE of the object is 4 times the amount it had when it had half the velocity. So it requires more energy the faster it goes...  thus more force to keep the same rate of acceleration.

KE = 1/2 MV^2        

Please hit me with -ve if I am wrong here - I am getting this from my head rather than looking it up. The force between 2 objects falls off via an inverse square law with distance.

F = G.M1M2 / Cr^2      where C here is something like 4pi x Epsilon(0) or something like that. G is gravitational constant.

NOOOOOOOO
All the exapmles about free-falling go from the assumption that the gravitational Force is constant!!!!! (the heights are too small)
 

If the object has the mass of 20kg then the gravitational force (during the fall) is constant = 20*9.8=196 Newton !!!
In other words, it can be said that the size of the pushing per any one second of falling is 196 Newtons.

15 minutes ago, DrP said:

I don't think they are identical because KE = 1/2 MV^2    -  so the faster you go the more energy required to accelerate. In the case of gravity the force increases as you get closer to the ground as force is directly proportional to the inverse of the separation distance SQUARED.

You must know this though - what am missing now?

 

NOOOOOOO please read previous answer ( the gravitational force DOES NOT increase as you get closer to the ground) because the heights are too small. 

Edited April 3, 2018 by OlegGorokhov

[Strange]

 

24 minutes ago, DrP said:

I don't think they are identical because KE = 1/2 MV^2    -  so the faster you go the more energy required to accelerate. In the case of gravity the force increases as you get closer to the ground as force is directly proportional to the inverse of the separation distance SQUARED.

In general, though, you can ignore the changing acceleration due to gravity (for small displacements). So the two are equivalent. 

If you want to take int account the increasing force of gravity then, yes, Karlson would need to also increase the applied force. But that is an unnecessary complication (when Oleg won't even answer basic questions).

29 minutes ago, OlegGorokhov said:

Now science ERRONEOUSLY THINKS ...

You keep saying this (in fact, all you do is repeat yourself).

Stop just telling us that current science is wrong. Whether or not current theory is wrong says nothing about the correctness of your idea. Only evidence can do that.

Show us how YOU calculate energy in your model and how this compares with measurements of energy. You know, do some science.

(If you just repeat exactly the same thing again, I suspect this thread will get closed. And that will be the end of this "very important theory")

[OlegGorokhov]

29 minutes ago, Strange said:

 

In general, though, you can ignore the changing acceleration due to gravity (for small displacements). So the two are equivalent. 

If you want to take int account the increasing force of gravity then, yes, Karlson would need to also increase the applied force. But that is an unnecessary complication (when Oleg won't even answer basic questions).

Let's take the fact the gravitational force is constant when free-falling (since the the heights are too small). Ask your teacher about it.

The Karlson's example is needed to understand the point.

It's important to understand that
The 2nd second of the movement of the stone (under the action of the pushing Carlson) and the 2nd second of free fall of any stone (under the action of the pushing gravitational force) are completely identical processes.

These two processes are completely identical, because in both cases:
1) at the beginning of the 2nd second the stone already has the velocity of 9.8m/s to move at this velocity by inertia (during the 2nd second);
2) as a result of pushing the outer Force (Carlson in the first case and the gravitational Force in the second case), the velocity of the stone increases by 9.8m/s (during the 2nd second);
3) the total dispacement (during the 2nd second) is: 14.7m = 9.8m (inertial displacement) + 4.9m (non-inertial displacement).

Now science ERRONEOUSLY THINKS that the Energy of the gravitational Force (Karlsson’s Energy) is spent on this inertial displacement of the stone (9.8m) during the 2nd second !

So do you agree with the fact that  Karlson’s Energy will not be spent on this displacement of the stone (9.8m), during the 2nd second?
Because this displacement (9.8m, during the 2nd second!) would be even without Karlson.

 

Edited April 3, 2018 by OlegGorokhov

[Strange]

Reported for tolling.

But please feel free to post exactly the same thing again.

[OlegGorokhov]

1 hour ago, Strange said:

Show us how YOU calculate energy in your model and how this compares with measurements of energy. You know, do some science.

(If you just repeat exactly the same thing again, I suspect this thread will get closed. And that will be the end of this "very important theory")

First of all understand that the current formula for Work and Energy is is erroneous.
Do it through the Karlson's example.

 

And then the question is
So do you agree with the fact that  Karlson’s Energy will not be spent on this displacement of the stone (9.8m), during the 2nd second?
Because this displacement (9.8m, during the 2nd second!) would be even without Karlson.

 

THE KARLSON'S EXAMPLE IS NEEDED TO UNDERSTAND ABOUT THE GRAVITATIONAL FORCE

So do you understand that the gravitational does not do Work moving the stone trough the 9.8m dispalcement (during the 2nd second)?

Because
by the beginning of the 1st second of the fall, the 20kg object has not yet accumulated the velocity v (= 0 m/s). Therefore, during the 1st second of the fall the object moves a total distance D (= 4.9 m) for only one reason:
- the object moves the displacemen D1 (= 4.9 m) as a result of the Work (Pushing) of the gravitational Force, and
- the object does not move the displacemen (D2) (the distance that is due to inertia), because the object has not yet accumulated any velocity (v).

D = D1 + D2 = 4.9m + 0m = 4.9m

By the beginning of the 2nd second of the fall, the 20kg object had already accumulated the velocity v (= 9.8 m/s). Therefore, during the 2nd second of the fall, this object moves the total displacement D (= 14.7 m) already for two reasons:
- the object moves the displacement D1 (= 4.9 m) as a result of the Work (Pushing) of the gravitational Force; and
- the object moves the displacement D2 (= 9.8 m) as a result of movement by inertia with a velocity v (= 9.8 m/s), i.e. without the Work (Pushing) of the gravitational Force.

D = D1 + D2 = 4.9m + 9.8m = 14.7m

So do you understand that the gravitational does not do Work moving the stone trough the 9.8m dispalcement (during the 2nd second)?

Edited April 3, 2018 by OlegGorokhov

[DrP]

59 minutes ago, OlegGorokhov said:

NOOOOOOOO
All the exapmles about free-falling go from the assumption that the gravitational Force is constant!!!!! (the heights are too small)
 

If the object has the mass of 20kg then the gravitational force (during the fall) is constant = 20*9.8=196 Newton !!!
In other words, it can be said that the size of the pushing per any one second of falling is 196 Newtons.

NOOOOOOO please read previous answer ( the gravitational force DOES NOT increase as you get closer to the ground) because the heights are too small. 

OK - Sorry - the FORCE is constant...  but the distance over which the force is applied for is greater - so more work is done. To double the speed 4 times the energy is required.   

How about this  -  if carleson was a motor using fuel - he would need to use more fuel each step to increase the constant acceleration. This can be measured by looking at the fuel intake increase on his journey. 

 

 

 

40 minutes ago, Strange said:

 

In general, though, you can ignore the changing acceleration due to gravity (for small displacements). So the two are equivalent. 

If you want to take int account the increasing force of gravity then, yes, Karlson would need to also increase the applied force. But that is an unnecessary complication (when Oleg won't even answer basic questions).

Thanks - maybe I got confused by the inverse square law (goes from centre of masses? - So free falling r is considered constant due to negligible change in r)   -  The KE is increasing exponentially though as V doubles KE quadruples yea? This is easily tested via fuel consumption.

[swansont]

1 hour ago, OlegGorokhov said:

The Karlson's example is needed to understand this point.

Are you going to answer my question or not?

[Strange]

18 minutes ago, OlegGorokhov said:

First of all understand that the current formula for Work and Energy is is erroneous.
Do it through the Karlson's example.

 

And then the question is
So do you agree with the fact that  Karlson’s Energy will not be spent on this displacement of the stone (9.8m), during the 2nd second?
Because this displacement (9.8m, during the 2nd second!) would be even without Karlson.

Why are you unable to show us how YOU calculate energy in your model and how this compares with measurements of energy?

We are not here to answer your questions. You are here to show that your model works. I don't understand why you are refusing to do that.

Edited April 3, 2018 by Strange

[OlegGorokhov]

54 minutes ago, swansont said:

Are you going to answer my question or not?

Your questions were 
No. That's not how this works. You answer my questions. You defend your claims.
You explain the situations where you think standard physics fails, and/or how your solution is required to come up with the correct answer.
I don't care about Karlson's energy. I just want to solve the problem. 
Solve the problem with your method. 
Explain why standard physics doesn't give you the right answer.
Answer the questions, as you promised to do. Don't just repeat yourself.

I did answer you to read again (and carefully) the the Karlsons example.

You repeat,
"I don't care about Karlson's energy. I just want to solve the problem. "


THE KARLSON'S EXAMPLE IS NEEDED TO UNDERSTAND ABOUT THE GRAVITATIONAL FORCE

So do you understand that the gravitational Force does not do Work moving the stone trough the 9.8m dispalcement (during the 2nd second)?

Because
by the beginning of the 1st second of the fall, the 20kg object has not yet accumulated the velocity v (= 0 m/s). Therefore, during the 1st second of the fall the object moves a total distance D (= 4.9 m) for only one reason:
- the object moves the displacemen D1 (= 4.9 m) as a result of the Work (Pushing) of the gravitational Force, and
- the object does not move the displacemen (D2) (the distance that is due to inertia), because the object has not yet accumulated any velocity (v).

D = D1 + D2 = 4.9m + 0m = 4.9m

By the beginning of the 2nd second of the fall, the 20kg object had already accumulated the velocity v (= 9.8 m/s). Therefore, during the 2nd second of the fall, this object moves the total displacement D (= 14.7 m) already for two reasons:
- the object moves the displacement D1 (= 4.9 m) as a result of the Work (Pushing) of the gravitational Force; and
- the object moves the displacement D2 (= 9.8 m) as a result of movement by inertia with a velocity v (= 9.8 m/s), i.e. without the Work (Pushing) of the gravitational Force.

D = D1 + D2 = 4.9m + 9.8m = 14.7m

So do you understand that the gravitational Force does not do Work moving the stone trough the 9.8m dispalcement (during the 2nd second)?
 

Edited April 3, 2018 by OlegGorokhov

[swansont]

2 hours ago, swansont said:

 Solve, using your method, for the kinetic energy of an object subject to a force F through a displacement d.

 

1 hour ago, swansont said:

How about solving: How far does an object drop under gravity in 2 seconds?

Last chance

[Strange]

13 minutes ago, OlegGorokhov said:

THE KARLSON'S EXAMPLE IS NEEDED TO UNDERSTAND ABOUT THE GRAVITATIONAL FORCE

...

So do you understand that the gravitational does not do Work moving the stone trough the 9.8m dispalcement (during the 2nd second)?

Lets just take it for granted that after so many repetitions that we understand what you are saying.

Now: use your model to make some quantitative testable predictions.

Or admit you can't. Either is fine.

Edited April 3, 2018 by Strange

[John Cuthber]

10 hours ago, OlegGorokhov said:

Energy (E) is the ability to push an object with a Force (F) for a time (t).

 

OK, I left a 1 KG weight on a shelf a year ago.
According to you, it now has (or has used)  52 times more energy than a 1kg weight I put on the same shelf a week ago.

 

Do you have any idea how little sense that makes?
 

[OlegGorokhov]

2 minutes ago, swansont said:

Solve, using your method, for the kinetic energy of an object subject to a force F through a displacement d.
How about solving: How far does an object drop under gravity in 2 seconds?

Did you read the 3rd link???

But I think it will be too difficult for you now.



First please answer the following question. (about -- How about solving: How far does an object drop under gravity in 2 seconds?)
Are you smart enough to read the text and the table

So your  question is
How about solving: How far does an object drop under gravity in 2 seconds?

The answer is
4.9m+14.7m= 19.6m 

Once again
4.9m (the displacement during 1st second) + 14.7m (the displacement during 2nd second) = 19.6m in 2 seconds

Capishe?
 

 

3 minutes ago, John Cuthber said:

OK, I left a 1 KG weight on a shelf a year ago.
According to you, it now has (or has used)  52 times more energy than a 1kg weight I put on the same shelf a week ago.

 

Do you have any idea how little sense that makes?
 

The answer will be in the Discovery#3 (about The New Law of Conservation of Energy.)
Discovery #3: (temporally hidden now).

If this new knowledge (that Energy of the table is spent without any movement of 20kg of the object) confuses you, just think about how many small particles are , for example, inside the iron table.These particles are pushing each other to stay together in form of the table.
These pushings are Work, and, of course, Energy is wasted on this Work.
Think about how much Energy is spent on these pushings every second, in order the table simply remains in the form of a firm table.
And, of course, extra Energy is spent on these pushings when there is the 20kg object on the table’s surface to prevent the particles from dispersing apart from each other and to prevent this 20kg object from going downwards (this is what the 20kg object is trying to do every moment).

[Strange]

5 minutes ago, OlegGorokhov said:

Did you read the 3rd link???

Now you have shown us your equation, please use it to compare the results to measured energy.

 

6 minutes ago, OlegGorokhov said:

Did you read the 3rd link???

Did you read the rules?

[OlegGorokhov]

6 minutes ago, John Cuthber said:

OK, I left a 1 KG weight on a shelf a year ago.
According to you, it now has (or has used)  52 times more energy than a 1kg weight I put on the same shelf a week ago.

 

Do you have any idea how little sense that makes?
 

 

The answer will be in the Discovery#3 (about The New Law of Conservation of Energy.)
Discovery #3: (temporally hidden now).

If this new knowledge (that Energy of the table is spent without any movement of 20kg of the object) confuses you, just think about how many small particles are , for example, inside the iron table.These particles are pushing each other to stay together in form of the table.
These pushings are Work, and, of course, Energy is wasted on this Work.
Think about how much Energy is spent on these pushings every second, in order the table simply remains in the form of a firm table.
And, of course, extra Energy is spent on these pushings when there is the 20kg object on the table’s surface to prevent the particles from dispersing apart from each other and to prevent this 20kg object from going downwards (this is what the 20kg object is trying to do every moment).

 

[Strange]

Just now, OlegGorokhov said:

The answer will be in the Discovery#3 (about The New Law of Conservation of Energy.)
Discovery #3: (temporally hidden now).

why not just tell us the answer? Stop playing games.

[John Cuthber]

11 minutes ago, OlegGorokhov said:

Did you read the 3rd link???

No.

I stopped as soon as I saw that you plainly don't make sense.

5 minutes ago, OlegGorokhov said:

These pushings are Work, and, of course, Energy is wasted on this Work.

Not according to the definition of work (as far as Physics is concerned).

 

I'm sure others have asked this

What experimental proof do you have that the current version of physics is wrong?
What evidence do you have that your version works better?

 

If you don't have clear answers to those questions you might as  well stop wasting bandwidth.

 

[OlegGorokhov]

8 minutes ago, Strange said:

why not just tell us the answer? Stop playing games.

Do not take on yourself too much

First get the Karlson's example (about 2 identical processes during the 2nd second ).

First get the fact that

The 2nd second of the movement of the stone (under the action of the pushing Carlson) and the 2nd second of free fall of any stone (under the action of the pushing gravitational force) are completely identical processes.

These two processes are completely identical, because in both cases:
1) at the beginning of the 2nd second the stone already has the velocity of 9.8m/s to move at this velocity by inertia (during the 2nd second);
2) as a result of pushing the outer Force (Carlson in the first case and the gravitational Force in the second case), the velocity of the stone increases by 9.8m/s (during the 2nd second);
3) the total dispacement (during the 2nd second) is: 14.7m = 9.8m (inertial displacement) + 4.9m (non-inertial displacement).

Now science ERRONEOUSLY THINKS that the Energy of the gravitational Force (Karlsson’s Energy) is spent on this inertial displacement of the stone (9.8m) during the 2nd second.

So NOW do you understand that the gravitational Force does not do Work moving the stone trough the 9.8m dispalcement (during the 2nd second)?

Edited April 3, 2018 by OlegGorokhov

[Strange]

3 minutes ago, OlegGorokhov said:

Do not take on yourself too much

First get the Karlson's example (about 2 identical processes during the 2nd second ).

Please stop repeating yourself and answer the question.

Quote

The 2nd second of the movement of the stone (under the action of the pushing Carlson) and the 2nd second of free fall of any stone (under the action of the pushing gravitational force) are completely identical processes.

Yes, we know that.

3 minutes ago, OlegGorokhov said:

Now science ERRONEOUSLY THINKS that the Energy of the gravitational Force (Karlsson’s Energy) is spent on this inertial displacement of the stone (9.8m) during the 2nd second.

Great.

Now use some mathematics to show the predictions of your model and compare the results with measurement.

Edited April 3, 2018 by Strange

[John Cuthber]

2 minutes ago, OlegGorokhov said:

So NOW do you understand that the gravitational Force does not do Work moving the stone trough the 9.8m dispalcement (during the 2nd second)?

No.

I don't.

 

If a stone falls further before it lands on someone's head then it will do moredamage.

To do that damage it needs energy- the energy needed to break bone etc.

If the additional energy doesn't come from gravity, where does it come from?

The only thing acting an a stone in free fall is gravity.

It is the only possible source of the additional energy.

And I therefore conclude that you are hopelessly wrong.

Let us know if anything changes.

 

[OlegGorokhov]

1 minute ago, Strange said:

Please stop repeating yourself and answer the question.

Great.

Now use some mathematics to show the predictions of your model and compare the results with measurement.

I repeat myself cos you DO NOT GET the elenentary things.
Just get it. Read it again. Read it again SLOWLY.

First understand the mistake in the current formula for Work and Energy!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
First get the Karlson's example and about the gravitational Force (about 2 identical processes during the 2nd second ).
Read the text again, again and again
 

[Phi for All]

15 minutes ago, OlegGorokhov said:

The answer will be in the Discovery#3 (about The New Law of Conservation of Energy.)
Discovery #3: (temporally hidden now).

!

Moderator Note

OlegGorokhov, you've been given three pages to explain why all the claims you're making are valid. The members have been extremely civil in the face of your constant repetition, and your ignoring of their simple questions to help you test your idea. You are NOT doing science by behaving this way. 

And the above shows that you have some non-science agenda to grandstand your idea. This is a science DISCUSSION site, we don't allow soapboxers who don't listen, we don't allow preachers who ignore questions. This thread is closed, and you are NOT allowed to open this subject again until you can answer ANY of the questions you've been asked.

 

1 minute ago, OlegGorokhov said:

I repeat myself cos you DO NOT GET the elenentary things.
Just get it. Read it again. Read it again SLOWLY.

First understand the mistake in the current formula for Work and Energy!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
First get the Karlson's example and about the gravitational Force (about 2 identical processes during the 2nd second ).
Read the text again, again and again

!

Moderator Note

I also want to say I'm pleased that the rest of the membership didn't sink to attacking the intellectual integrity of another poster the way Oleg is doing here. Thread closed.

 

[swansont]

16 hours ago, OlegGorokhov said:

Did you read the 3rd link???

No, I did not.You have been told to discuss the material here. Nobody should have to go through a link to get information regarding your thesis.  

Quote

 
 

A 2N force (1 kg at 2m/s^2) through a distance of 2 meters gives a KE magnitude of 2*sqrt2, while in conventional physics the answer would be 4. So that's a problem.

But what if my 2N force was for a 2 kg object and 1m/s^2 of acceleration? The answer changes! That's another problem.

Yet another problem is that you have units of kg-m/s, which are not units of energy

!

Moderator Note

I will add to Phi's modnote that because the thesis is demonstrably false, as shown above, do not bring this up again.