re: energy producing experiments

[Kator01]

Delburt


as I said im my last comment about your accelleration-calculation " That is what I am not sure about"


So I was looking up an old german Physics-Textbook Grimsehl-Tomaschek, dated 1940. In a chapter about the equivalence-condition
of rotating masses I found an experimental setup in which it is demonstrated that the accelleration of a falling mass which
rotates two masses ( see pic Mass-Equiv_02.jpg)  at both ends of a horizontal mounted bar stays always the same if the value of the moment of inertia ( m*r²)
stays the same for different symetrical positions of the two masses on the bar. In order for (m*r²) to be constant, the masses have to be adapted if their
radial posistion changes.


If you have 2 x 1 kg at r = 0.8 m, moment of intertia I = 2(1kg * 0.64 m²)-> I = 1.28 kg m²


Now you want to move the two masses to r = 0.4m then for I=1.28 kgm² to remain constant ( 1.28) the two masses have to be increased ( I use here the
term replacement-mass m') to


                       m'= 1.28 / 0.4² = 1.28/ 0.16 = 8kg in total, 4 kg at each side,


to fullfill the condition that the accelleration of the falling mass stays the same as with 2 * 1 kg masses at r = 0.8m.


The calculations made in this old book are based on the cgs-System ( centimeter-gramm-second )
and only with this system the formula for the resulting accelleration b of weight G is true :


                       G (gr) * 981 (dyn) = b * (m * r² + G) ( see pic Equiv-Mass_03.jpg) 
 
The formula is correct only for r = 1 cm, as the value of r² = 1 ....so that it is save to assume that the replacment-mass m'
experieneces the same accelleration as G  ( falling weigth )


         b = G (gr) * 981 (dyn) / (m (gr) + G (gr))                            ( Formula Nr. 1 )


......but what would be the new total replacement-mass m' on both sides of the bar at r = 1 cm ?


The moment of Inertia for the two 1kg masses ( in gramm ) at r=80 cm


      I = 2000 gr * (80cm)² = 2000 * 6400 = 12.800.000 gr cm²


In order to have the same moment of Inertia at r = 1cm the new mass m' = 12.800.000 / 1
that is two masses of 6.400.000 gr ( 6.400 kg) at each side.


According the Formula Nr.1 above, falling mass = 40 kg = 40.000 gr


                b = 40 000 g * 981 dyn / (12.800.00gr + 40.000gr) = 39.240.000 /12.840.000
               
                b = 3.05 cm/ s² = 0.0305 m/s²   !!!!
______________________________________________________________________________________




Now in order toget an idea the accekeratio of G with an unchanged replacement-mass m' at r=2 cm ( although this not correct )
we end up with b = 0.121 m/s²


From here you calculate the final velocity of 40kg falling down 1 meter : V = Square root of (2 * 0.121 * 1) = 0.492 m/s


Since the axis-radius is 2 cm the circumference of the axix = 0.126 m
With the velocity 0f 0.492 m/s / 0.126  you get 3.9 revolutions per second.


The cirumference at which the two 1kg masses rotate is : 2 * Pi * 0.8 = 5 m


With 3.9 Revolutions / s the rim-velocity of the two 1 kg masses is 19.5 m/s


Kinetic energy Ekin = 2kg/2 * (19.5)² m²/s² = 380.25 Joule


Input was : m * g * h = 40kg * 9.81 m/s² * 1m = 392,4 Joule


The missing 12 Joule are accounted to the falsy used r = 2cm with the unchanges m' in the Formula N.1


So there is no gain in this system. That is the reason why we dont have overunity-bicycles




Mike


Edit: Sorry for the big rotated pics. I tried to remove and replace them ..but to no avail

[Delburt Phend]

Surly you do not believe that 1 kg at 80 cm is balance by 6,400 kg at 1 cm.

Shortly after (although this is not correct) a number comes up from nowhere; .121 m/sec/sec and then this mystery number is used for calculation.

My experiments are real.

[Delburt Phend]

A lever arm can't move up until it is first balanced.

The 80 to 1 lever arm, according to your math, can't be moved up until 6,400 kg are placed on the opposite short (1 cm) side. 

And: according to your math, then the long side of a 40 to 1 lever arm can't be moved up until 1,600 kg are placed on the opposite short (1 cm) side. But the one kilogram on the 40 cm side balances with 40 kg on the 1 cm side; and you then add another 40 kg for acceleration.

But (for you) this 80 kg is still (1600 – 80) 1520 kg short of being balanced. If your rule worked, then the 1 kg at 40 cm will accelerate the 80 kg; not the other way around. For your math the 80 kg at 1 cm * 1 cm; should be overpowered by 1 kg * 40 cm* 40 cm.

----
Experiments collect data to establish or support a concept; so an experiment will collect data points. The concept is; If a longer lever arm with a proportional less mass is harder to rotate, then it will be harder to rotate. For example: if one side of a pivot point is 40 kg at 1 cm and the other side is 1 kg at 40 cm; then according to the mr² theory the long side is 40 time harder to rotate than the short side.

The Laws of Levers says that the two sides are equally hard to rotate. So, let's construct an experiment that separates the true from the false.

If you apply the same force to two 40 kg masses at 1 cm on both sides; and to two 1 kg masses at 40 cm on both sides, then the harder to rotate side will take longer to rotate, or it will have a smaller acceleration.

The Force is held to be the same by using the same suspended mass at the same location. In the 40 to 1 we used an extra 40 kg on the short side.

You find data points to determine if the acceleration is the same. A 26 mm gate trip at the same times after the two Atwood's (two 40 kg at 1 cm, and two 1 kg at 40 cm) has moved the same distance (real numbers .0628 and .0631 sec for one of many set ups).  Having the same final velocity after moving the same distance assures us that the different Atwood's have the same acceleration.

Having the same F divided by the same 'a' for both Atwood's (40 kg at 1 cm both side; and 1 kg at 40 cm both sides) assures us that the inertia 'm' has to be the same also.

Real numbers for a certain Atwood's are:  two 20.7 kg at a 1.03 cm r; compared to two .8628 kg at 24.13 cm. Both took 3.3 sec; in another set up. They had the same F and the same 'a' and the same inertia 'm'.

[Delburt Phend]

I caught myself wondering how science made this mr² mistake in the first place. And then I remembered: it is all about the protection of ½ mv² (probably mv² at the time); if 1 kg can be accelerated to 2 m/sec as easily as 2 kg can be accelerated to 1 m/sec then The Law of Conservation of Energy goes kaput. And that is the sacred cow of physics.   

[Kator01]

Delburt

The 80 to 1 lever arm, according to your math, can't be moved up until 6,400 kg are placed on the opposite short (1 cm) side. [/size]And: according to your math, then the long side of a 40 to 1 lever arm can't be moved up until 1,600 kg are placed on the opposite short (1 cm) side. But the one kilogram on the 40 cm side balances with 40 kg on the 1 cm side; and you then add another 40 kg for acceleration.

[/size]
No, this is not what is meant please reread my description.[/size]

[/size]
We have to distinguish between a static situation ( Nm)  and a dynamic one-> accelleration..--> mr² ..resistance to rotary accelleration.[/size]

[/size]
"By the way: this is not "my math",  I just try to understand the basics and s[/size]ince you changed to a lever-system ( away form the Cyl & sphers) in order to find an easier way ( compared to the Cyl. & Sph) I was forced to dive into the depth of what was known to this subject back then.[/size]

[/size]
I cannot and will not argue whether this is true of not but I understood the principle of this experiment. [/size]

[/size]
One thing is sure that these guys back then did extensive and very elaborated experiments. I spent a few days searching the internet ...didnt find anything and then looked the subject up in my old book. I have no time to translate all this to english but the experiment was basis in university
teachings and sure enough they demonstrated this in front of students.
If I had the time and the means I would build this experiment.


I am a pragmatic person, i.e. I build, experiment and analyse...like you to. This I did with the Cylinder & Spheres experiment. Something went wrong
with a brass-bearing and I have to re-engineer the setup. However I could see and film that momentum is conserved not energy. I will continue with this small wheel setup and build it according to sound mechanical means and do not let myself distracted by other setups because I have not reached the core-function of the Cyl & Sphere system.


I agree: They do everything to control attention to protect their narrative.

Levine on moment of inertia

https://www.youtube.com/watch?v=lvfzdibrUFA

Mike