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### Author Topic: re: energy producing experiments  (Read 147153 times)

#### kolbacict

• Hero Member
• Posts: 1418
##### Re: re: energy producing experiments
« Reply #240 on: November 15, 2021, 06:41:26 PM »
is that talk about the same?

#### Delburt Phend

• Full Member
• Posts: 223
##### Re: re: energy producing experiments
« Reply #241 on: November 16, 2021, 05:45:57 PM »
The Administrator is welcome to remove the kolbacict post. I thought I would post again so first timers see something that is on subject.

Look at the last statement:  My point is: one way to make energy is to combine a modified Atwood’s with a rotating balanced beam.

You start with a modified Atwood’s. You have a 4 kg block on dry ice. The block is tied to a string that is draped over a pulley that has a 1 kg mass suspended from the other end.

The acceleration is 1 kg / 5 kg * 9.81 m/sec² = 1.962 m/sec²

The one kilogram is allowed to accelerate toward the floor for one meter. This also means that the 4 kilograms on the dry ice will accelerate for one meter at a rate of 1.962 m/sec.

After the one kilogram has dropped one meter you have 4 kilograms moving horizontally on dry ice at 1.9809 m/sec.

And you have 1 kg moving down at 1.9809 m/sec.

We then change the direction of the one kilogram and have it moving on a horizontal plane. Let’s direct the one kilogram so that it is moving in the opposite direction of the 4 kilograms.

So we have a 4 kg mass moving east on a frictionless plane. About a meter under it we have a 1 kilograms mass moving west.

At this point we connect the two moving masses to each end of a massless beam a meter long.

On this beam the two masses will rotate about their center of mass but the ‘tangent velocity’ of each mass will not remain the same. The total ‘tangent’ momentum will remain the same but the individual velocities will not.

The momentum of the one kilogram after it has dropped one meter is 1.98 m/sec * 1 kg = 1.98 units of momentum. The ‘tangent’ momentum of the one kilogram as it enters the rotation on the beam is 1.98 kg m/sec.

The momentum of the 4 kg is (4 kg * 1.98 m/sec = 7.9236) 7.9236 kg m/sec.

The total momentum is 7.9236 kg m/sec + 1.98 kg m/sec = 9.9 kg m/sec.

When an object is rotating about its center of mass it is also rotating about its center of ‘tangent’ momentum. So the 1 kg has half the momentum so it will be moving 4.95 m/sec.

The total momentum of the beam remained 9.9 kg m/sec.

At 4.95 m/sec the one kilogram can rise 1.25 m.

If the 1 kg is given all the momentum, as in the cylinder and spheres, it will rise; d = ½ v²/a;  5 m

So we start with a modified Atwood’s machine and then we place those masses in a massless beam. We then arrange for a cylinder and spheres event and we get a 500 % increase in energy.

I am going to point out your errors as they occur.

A line does not have to be straight; that is the whole point of curvilinear and rectilinear. You have at least two types of lines; curved and straight. You can’t properly look at Newtonian physic and have an improper definition of line. Your side want to set up a difference set of rule for mass moving in a circle; but Newtonian physics is about mass that is often moving in a circle.

Florens quote:  “It will also have to spin at a larger radius, so, by the law of levers it will decrease its speed, because v/r is constant.”

Wrong: not if you are talking linear v. This is where your misconception of linear messes you up every time.  Linear speed does not care one iota about r. It is the same as changing direction; you can change direction with a long pendulum or a short pendulum. You are not going to change v by using a long or short r.

#### sm0ky2

• Hero Member
• Posts: 3948
##### Re: re: energy producing experiments
« Reply #242 on: November 16, 2021, 11:07:30 PM »
The common misconception with an Atwood cylinder
is that they think they observe a gain in energy
due to an observation made from the velocity of the
rolling cylinder.

In actuality this has more to do with aerodynamics of rolling cylinders.
and why the cylinder-plane flies.

Momentum is conserved, even in an Atwood

#### Delburt Phend

• Full Member
• Posts: 223
##### Re: re: energy producing experiments
« Reply #243 on: November 17, 2021, 12:14:46 AM »
I did not know that an Atwood's (machine) had anything to do with a cylinder. Honestly what are you talking about?

#### sm0ky2

• Hero Member
• Posts: 3948
##### Re: re: energy producing experiments
« Reply #244 on: November 17, 2021, 02:24:33 AM »
Oh…  ok so history for the chronologically gifted

Well some time after the death of Atwood,
his knowledge was advanced as far as science could
(a pully is generally cylindrical)

The most advanced form took place in space
in microgravity, using a cylinder and 2 spheres
on strings, that wound around the cylinder (pulley)

Different masses were tested etc.

#### sm0ky2

• Hero Member
• Posts: 3948
##### Re: re: energy producing experiments
« Reply #245 on: November 17, 2021, 02:26:15 AM »
The results confirmed that momentum is always conserved

#### sm0ky2

• Hero Member
• Posts: 3948
##### Re: re: energy producing experiments
« Reply #246 on: November 17, 2021, 02:28:55 AM »
What i am talking about is that you can use
the momentum of the large weight to
throw the smaller weight more than it ‘should’
but you are missing the point that you putt that
into the larger weight to begin with

#### kolbacict

• Hero Member
• Posts: 1418
##### Re: re: energy producing experiments
« Reply #247 on: November 17, 2021, 08:35:29 AM »
Quote
Re: re: energy producing experiments
« Reply #241 on: November 16, 2021, 05:45:57 PM
The Administrator is welcome to remove the kolbacict post.
You are some kind of angry ...

#### Delburt Phend

• Full Member
• Posts: 223
##### Re: re: energy producing experiments
« Reply #248 on: November 17, 2021, 09:06:51 PM »
I know this is a recent repeat but I want to cover the first paragraph.

And the ‘MIT Atwood’s machine’ has an input momentum of .0443 units and an output of .4624 units of momentum.    .01 kg * 4.429 m/sec            1.11 kg * .41666 m/sec

If this MIT Atwood’s was a modified Atwood’s you would have 1.100 kg moving .4166 m/sec (on a plane) and, one meter away, there would also be .01 kg moving .4166 m/sec. If these two masses were placed on the ends of a balanced beam; and the beam were rotating about its center of mass, then the 1.100 kg would be moving .2102 m/sec and the .01 kg would be moving 23.135 m/sec. At 23.135 m/sec the .01 kg will rise 27.255 m.

The .01 kg will rise 27.255 m and it was dropped 1 meter.

This 27.25 times more energy is done without transfer all the motion to the small mass. If you transferred all the motion to the small mass (as in the cylinder and spheres) the energy increase would be 110 times (ideal).

"And the ‘MIT Atwood’s machine’ has an input momentum of .0443 units and an output of .4624 units of momentum.    .01 kg * 4.429 m/sec    1.11 kg * .41666 m/sec"

The input momentum (.0443) is the mass times the velocity achieved by dropping a .01 kg mass 1 meter.

Any mass dropped one meter will develop a velocity of 4.429 m/sec. Therefore the momentum needed to throw an object up one meter is 4.429 units of momentum for each kilogram. Ten grams is .01 kg so it only needs .0443 units of momentum in order to rise 1 meter.

The .01 kg was dropped one meter; so to return the system to it original starting configuration .0443 kg m/sec is all that is needed.

The MIT Atwood’s covers one meter in 4.80 seconds for and average speed of .20833 m/sec, and a final speed of .41666 m/sec. The center of mass of the 1.100 kg does not move. But the 1.100 kg is moving .4166 m/sec at the end of the drop of the .01 kg mass.

So the output momentum is 1.110 kg * .41666 m/sec = .4625 kg m/sec

This is an increase of .4625 - .0443 = .4182 kg m/sec

We can restart the MIT Atwood’s ten times. That is 1000% of the energy we put in.

smOky2: Could you post links or an address for the Atwood’s experiment conducted in space? Thanks

#### sm0ky2

• Hero Member
• Posts: 3948
##### Re: re: energy producing experiments
« Reply #249 on: November 18, 2021, 01:12:14 AM »
m1*g*h = 1/2 m1 * v(final)^2 + 1/2 m2 * v(final)^2 + m2 * g * h
———————————————————————————————————

m1*g*h - m2*g*h = 1/2(m1+m2)*v(final)^2
————————————————————————-
(m1-m2)gh=1/2(m1+m2)*v(final)^2
———————————————————-

2[ (m1-m2)g/(m1+m2)]h=v(final)^2

it doesnt matter when or where you stop it
when or where you transfer the energy between the 2 masses

#### Delburt Phend

• Full Member
• Posts: 223
##### Re: re: energy producing experiments
« Reply #250 on: November 18, 2021, 02:27:23 AM »
You might explain what you are trying to say.

I assume you are saying that the energy of the .01 kg mass dropped 1 meter (v = 4.429 m/sec) is equal the energy of the 1.110 kg Atwood’s moving .41666 m/sec: which is quite correct.  ½ m v²

I am not proposing that you give the energy of the 1.110 kg to the .01 kg I am proposing that you give the momentum of the 1.110 kg to the .01 kg

The 1.110 kg has .4625 units of momentum; if you give all of that momentum to the .01 kg it will be moving .01 kg * v = .4625 kg m/sec.   v = .4625 kg m/sec / .01 kg = 46.25 m/sec.

At 46.25 m/sec it will rise 109 meters. It was dropped 1 m.

Cylinder and spheres machine give the motion of a massive cylinder to much smaller mass.

#### sm0ky2

• Hero Member
• Posts: 3948
##### Re: re: energy producing experiments
« Reply #251 on: November 18, 2021, 02:55:43 AM »
Their momenti are not separate, but part of one system.

In 238 years you are not the first person to walk this path

#### sm0ky2

• Hero Member
• Posts: 3948
##### Re: re: energy producing experiments
« Reply #252 on: November 18, 2021, 03:12:02 AM »
v = (sqrt)[19.6(1.11-0.1)/(1.11+0.1)]
———————————————————
v = 4.04479055 m/s

#### Delburt Phend

• Full Member
• Posts: 223
##### Re: re: energy producing experiments
« Reply #253 on: November 18, 2021, 04:11:16 AM »
Name the others that have walked this path.

To transfer momentum from a large mass to a small mass is incredibly simple.

But; name the others that have walked this path.

v = (sqrt) [19.6  .01 kg /1.110 kg= .4202 m/sec   ideal

The experimental value was very close  .41666 m/sec

#### sm0ky2

• Hero Member
• Posts: 3948
##### Re: re: energy producing experiments
« Reply #254 on: November 23, 2021, 02:18:00 AM »
The momenti are one and the same object, they are linked.
It is not two separate momenti (one for each mass),
But rather one single quantity, from both interconnected masses in motion.
If the large mass suddenly becomes disconnected (no transfer of momentum),
the smaller mass is accelerating downwards at 9.8m/s/s
its’ upward momentum being instantly disconnected as well.
Cut the string

If a scenario were present such as an impact lever at the very bottom,
point of maximum velocity, such that the large mass hit and transferred
100% of its’ momentum into an upward thrust, sending the smaller mass skyward.
And you were to catch it at its uppermost peak::::: even with 0 losses from wind or anything
(pay attention here!!)

The Potential Energy in the small mass:
E=m[small] * 9.8m/s/s * height = m[large]*9.8m/s/s* initial height

Do the tests yourself

This has been done, literally, millions of times