Language:
To browser these website, it's necessary to store cookies on your computer.
The cookies contain no personal information, they are required for program control.
the storage of cookies while browsing this website, on Login and Register.
 Storing Cookies (See : http://ec.europa.eu/ipg/basics/legal/cookies/index_en.htm ) help us to bring you our services at overunity.com . If you use this website and our services you declare yourself okay with using cookies .More Infos here: https://overunity.com/5553/privacy-policy/If you do not agree with storing cookies, please LEAVE this website now. From the 25th of May 2018, every existing user has to accept the GDPR agreement at first login. If a user is unwilling to accept the GDPR, he should email us and request to erase his account. Many thanks for your understanding

Custom Search

### Author Topic: re: energy producing experiments  (Read 134826 times)

#### lancaIV

• elite_member
• Hero Member
• Posts: 5165
##### Re: re: energy producing experiments
« Reply #330 on: March 20, 2022, 11:18:30 AM »
IancaIV, what point are you trying to make? Let us look at context:
Using online calculators:
https://www.omnicalculator.com/physics/free-fall
https://www.calculatorsoup.com/calculators/physics/gravitational-potential.php
https://www.calculatorsoup.com/calculators/physics/kinetic.php

The 140KG falls from 0.25m to achieve this speed. (343 J)
1kg can rise to 35m at 26.24 kgm/s, which also costs 344 J
There is a difference in momentum, bu where is the energy gain?
So unless I am missing something, it takes 51,200 J to "place" 310 units of momentum on 1kg. How did we magically transfer 310 units of momentum from a 140kg at 2.21m/s mass to a 1kg mass at 310m/s?

" 310 units of momentum and you only need 26.24 units " :  310/26,24 = ?
https://overunity.com/1763/12-times-more-output-than-input-dual-mechanical-oscillation-system/msg564829/#new

magica,mysteria,Magister,Meister,Master             M.S.A. :   Master Science Administration ~ kybernetes

I can,if You want to introduce Yourself in deeper hyper/deeper physics ,in"Quantum-Mechanics"-Logics explain ,similar ,
but animistic language,less " swollen ":

String to Point-Matrix
https://en.wikipedia.org/wiki/Urelement
C.Colombo : chicken or egg                      hyper/deeper physics Dualism:   spin or point

or,going 50 years back to its/his,Otto Stein,first publication date :

german,deutscher,Sprache/language : international -multilingual -traduction available doing ?
https://www.yumpu.com/de/document/view/20184764/otto-stein-die-zukunft-der-technik-pdf

Sincerely and a sinfull Sunday wishing
OCWL

#### Delburt Phend

• Full Member
• Posts: 222
##### Re: re: energy producing experiments
« Reply #331 on: April 25, 2022, 06:33:06 PM »
What do you think of Destin's evaluation?

#### Kator01

• Hero Member
• Posts: 894
##### Re: re: energy producing experiments
« Reply #332 on: May 03, 2022, 12:41:03 AM »
Delburt
Excellent find.You are a keen observer. I am amazed of the different possibilities to fully transfer momentum of of a big mass to a much smaller one.
Lets calculate:
Mass of the knot including the fluffy part about 2 gr = 0.002 kg
Minimum Velocity according to Wikipedia: v = 342.2 m/sec at T = 20  degree Celsius

E_kin = 1/2 * 0.002 kg * 117 100.2 m_exp2/sec_exp2 = 117.1 Joule.........thats a lot

Remember: The fastest Pitcher can throw a ball 105.5 mph which is about 47 m per second and if you look at the ease with which this woman
throws the whip....

https://sportsnaut.com/fastest-pitch-ever-mlb-history/

There is one peculiar thing: as the wave rolles down to the end it stops its wave-like movement about 10 cm before the knot as then
the knot is accellerated to full speed like the head of a serpent.
Why that ?
Question : would the sound barrier be reached if the wave would roll down to the end ?
There is a small element in the construction of the last part of the whip which makes it work and these "smart three engineers have not figured it out,
they fully missed a key element of the "faster that sound-speed-function" in their experiment.
I will not reveal it...its a good excercise for sharpening ones observation-skills.

I call it "The serpents bite"

Mike_G

#### Delburt Phend

• Full Member
• Posts: 222
##### Re: re: energy producing experiments
« Reply #333 on: May 03, 2022, 03:53:38 AM »
There is one point in the video (smarter every day) where they say that the sound barrier is broken before the end; that might put the mass at around 20 grams. Also think of the air resistance you would get at these speeds.

#### Kator01

• Hero Member
• Posts: 894
##### Re: re: energy producing experiments
« Reply #334 on: May 03, 2022, 11:41:08 PM »
Delburt,

everthing is normal, see here:

https://de.wikipedia.org/wiki/%C3%9Cberschallflug#/media/Datei:Shockwave_pattern_around_a_T-38C_observed_with_Background-Oriented_Schlieren_photography_(1).jpg

and the compare to first and second picture

Now the third pic : what does this woman hold between their fingers ?
with the fourth pic : what does the corde look like shortly before the bang

Mike_G

#### Kator01

• Hero Member
• Posts: 894
##### Re: re: energy producing experiments
« Reply #335 on: May 04, 2022, 12:08:54 AM »
also I noticed recurring discussions whether formulas "moment of inertia" or "momentum" should be used . We dont need any further discussion on this as moment of inertia in the cylinder & the spheres experiment is not involved for any further calculations as the spheres - after getting all the momentum -  are freely rotating and do not create any torque via a spoke to the axis.

Its just simple as that. I have built a simple setup with a small flywheel from an old record player ( eighties) and can confirm the full-stop of the wheel if
the small mass is out at 90 degress. I used just one small mass. It will take me a longer time to build the complete setup as I need the assistance of a mechanic  which has his workshop 25 mile away

Mike_G

#### Delburt Phend

• Full Member
• Posts: 222
##### Re: re: energy producing experiments
« Reply #336 on: May 04, 2022, 01:39:51 AM »
At about 2:36 they talk about going supersonics before full extension; so a larger portion of the whip (in addition to the feathered end) must be moving at about the same speed.

Backyard scientist also says they went supersonic with rubber bands and a string. I am still working on that one.

#### Delburt Phend

• Full Member
• Posts: 222
##### Re: re: energy producing experiments
« Reply #337 on: August 24, 2022, 04:18:39 AM »
Laws of levers proves you can make energy in the lab.

Let’s apply a force to the lever arm at the same radius as the accelerated mass. Let’s use numbers to keep this clear. The accelerated mass is on a lever arm of .1 m radius; and the force is applied on a lever arm of .1 m radius.

The mass being accelerated is more difficult to accelerate, or more easily accelerated, in accordance with the Laws of Levers.    Fr

If the two lever arms are the same .1 m /.1 m them the mechanical advantage of the accelerated mass is 1. In this case you have a direct F = ma situation. If you apply 4 newtons of force to a 1 kg mass it will be moving 4 m/sec after one second. The energy is 8 joules

If you move the mass to .2 m radius; then the accelerated mass has a mechanical advantage of .2 m / .1 m = 2. To keep the torque the same at the .1 m position of the applied force you will have to reduce the mass at .2 m to .5 kg. The torque at the applied force location (.1 m) is the same so you will still get an F = ma acceleration: except now you will get .5 kg moving 8 m/sec.

If you move the mass to 4 m radius; then the accelerated mass has a mechanical advantage of 4 m / .1 m = 40. To keep the torque the same at the .1 m position of the applied force you will have to reduce the mass at 4 m to .025 kg. The torque at the applied force location (.1 m) is the same so you will still get an F = ma acceleration: except now you will get .025 kg moving 160 m/sec. The energy is 320 joules

Remember the second Law is “The change of momentum is proportional to the applied Force.” Note that all these rotating masses have the same momentum.

#### Kator01

• Hero Member
• Posts: 894
##### Re: re: energy producing experiments
« Reply #338 on: August 28, 2022, 02:36:08 AM »
Delburt

are you talking about a seesaw setup ? Mass on one side...force beeing applied on the other ?

Mike

#### Delburt Phend

• Full Member
• Posts: 222
##### Re: re: energy producing experiments
« Reply #339 on: September 02, 2022, 05:18:03 PM »
Consider a seesaw with 40 kg on one side and 1 kg on the other side. To be balanced the one kilogram must be 40 times further from the point of rotation than the radius of rotation of the 40 kilograms.

An extra force could be applied at the same radius as the 40 kg.

First: let us focus on the radius and rotation of the 40 kg. Let’s put the radius at 2 cm. So we have 40 kg at 2 cm on one side. We could also put 40 kg at 2 cm on the outer side of this balanced beam. There would be no rotation because we have no unbalanced force.

When we put an extra 40 kg at 2 cm on one side we will get acceleration. We have (40 kg + 40 kg+ 40 kg) 120 kg accelerated by 40 kg. This is 1/3 standard acceleration; 9.81 m/sec/sec / 3 = 3.27 m/sec/sec. We know this to be true from Atwood’s machine experiments.

Now let us switch out a 40 kg mass on both sides for a 1 kg mass at (2 cm * 40 = 80 cm) 80 cm. The two 1 kg masses at 80 cm are the same rotational resistance as the two 40 kg masses at 2 cm, because they are at forty times the radial distance.

Because the resistance remained the same the acceleration will also remain the same; 3.27 m/sec/sec for the remaining extra 40 kg at 2 cm. But the two one kilogram masses are moving 40 times faster than the 40 kg drive mass. At forty times faster they have forty times the energy. ½ * 40 kg *(2.557 m/sec)² = 130.76 joules     ½ * 1 kg * (102.28 m/sec)² = 5230.6 J

So let’s change this into a practical arrangement. The 2 cm radius could be a 4 cm shaft that is attached to a 160 cm beam. On the end of each side of the beam are 1 kg masses. The shaft can be accelerated by a string that is wrapped around the shaft; the string is suspending a 40 kg mass from its end. This would be 3.27 m/sec/sec acceleration at the surface of the shaft.

After a one meter drop of the 40 kg; the 40 kg will be moving (1 m = ½ v²/ 3.27 m/sec/sec) 2.557 m/sec. But the two 1 kg masses will be moving 102.28 m/sec. Their kinetic energy would be KE = ½ * 1 kg * (102.28 m/sec)² = 5230.6 Joules.

The input energy is 40 kg at 1 meter of height which is (40 kg * 9.81 N/kg) * 1 m = 394.4 joules. This is the force produced by the drive mass of 40 kg that was dropped 1 meter.

The output energy is 5230.6 joules * 2 + (½ *40 kg * 2.557 m/sec * 2.557 m/sec) = 10,591.96 Joules.

There were 394.4 joules put in: and 10,591.96 joules came out.

#### Tarsier_79

• Full Member
• Posts: 117
##### Re: re: energy producing experiments
« Reply #340 on: September 03, 2022, 10:50:38 AM »
Quote
The two 1 kg masses at 80 cm are the same rotational resistance as the two 40 kg masses at 2 cm, because they are at forty times the radial distance.

Except they are not.

#### Delburt Phend

• Full Member
• Posts: 222
##### Re: re: energy producing experiments
« Reply #341 on: September 03, 2022, 02:55:01 PM »
You can put the balanced masses at any distance you want. I have not used so large a radial difference (2 cm to 80 cm) but I have confirmed the principal with experiments, with less dramatic differences.

Are you speaking from experimental knowledge; and if so, please sight the experiment.

Fundamentally what you are saying is that in takes more force to rotate the long side of a balanced tube or beam; than the force needed to rotate the short side. Please explain how that might be possible as it is a ridged object. And how were you convinced of this without experiments?

#### Tarsier_79

• Full Member
• Posts: 117
##### Re: re: energy producing experiments
« Reply #342 on: September 03, 2022, 09:29:51 PM »
The inertia of a point mass in rotation is I=mr2
The two 1 kg masses at 80 cm will be much harder to accelerate than the two 40 kg masses at 2 cm.

The simple proof is the period of a pendulum. The formula is T2=L/g. Note there is also a squared relationship. The period of a pendulum is derived from its inertia and acceleration. The squared (or square root) part of the formula is from inertial math.

Quote
So when um, when you double the length of the pendulum, you see that the period increases by a factor of square root of two.

#### Delburt Phend

• Full Member
• Posts: 222
##### Re: re: energy producing experiments
« Reply #343 on: September 03, 2022, 10:31:31 PM »
If your mr² is correct, then a pendulum bob of 2L would be 4 times harder to move than a bob at 1 L; clearly this is a false statement. The inertia of a bob is unchanged no matter what the length of the pendulum. The bob accelerates itself it is not under torque from the point of rotation.

The period of a pendulum has to do with the length of the swing and the sine of the angle; this is not what is being discussed.

MIT Physics Demo -- Center of Mass Trajectory

Rotation about the center of mass requires that the linear Newtonian momentum (mv) on one side is equal to the linear momentum on the other side. It is not about angular momentum conservation or energy conservation. The energy of one side is not equal to the energy of the other side and the angular momentum of one side is not equal to the angular momentum of the other side.

If a 40 kg mass is attached by a long tube to a one kilogram mass on the other end, and it is flipped in the air, it will rotate about its center of mass.  And 40 kg * r = 1 kg * 40 r.

Explanation of the universe must correspond with what the universe does. And the universe does Laws of Levers and the linear momentum (mv) of one side is equal to the linear Newtonian momentum of the other side.

Do the experiments what do the experiments tell you; this is only worth about 10 trillion dollars.

#### Tarsier_79

• Full Member
• Posts: 117
##### Re: re: energy producing experiments
« Reply #344 on: September 04, 2022, 12:49:45 AM »
Quote
If your mr² is correct, then a pendulum bob of 2L would be 4 times harder to move than a bob at 1 L; clearly this is a false statement.

It is 4 x harder to move.

Quote
The inertia of a bob is unchanged no matter what the length of the pendulum. The bob accelerates itself it is not under torque from the point of rotation.
The reason the pendulum length is not a linear relationship to the period is the same reason inertia doesn't have a linear relationship with mass/radius. A pendulum is the same as your lever. Instead of looking with tunnel vision at the problem, consider the similarities. Gravity provides a constant acceleration F=MA
If I double the length of a pendulum/lever-weight, Gravity doubles the torque on the system (leverage), but it doesn't take the same time to rotate. (because it is squared)

Quote
The period of a pendulum has to do with the length of the swing and the sine of the angle; this is not what is being discussed.

That is the funny thing with Energy math. Everything relates to each-other, as it should.

The period of a pendulum, gravity accelerates a mass around a pivot point. Your see-saw 2 masses are being accelerated around a pivot point. Regardless of where you apply torque: to the center of rotation, or the furthest mass from the center, the inertia increases exponentially.