energy producing experiments

[Tarsier79]

Hi Ruggerodk

The large radii spinning mass has a better wind up rate than a small radii: 50% vs 33%.
So it seems that a spinning weight at larger radii is better suited to collect and save energy...
Why is that?

Considering just the driver at different radii: at a smaller radius, the torque the mass applies, and the time it takes to do so vs friction probably plays an important part here.

1. If it takes longer to unwind, the friction partakes for longer, sapping more energy.

2. At a smaller radius, your torque vs friction ratio is different, so I would expect a poorer drive effect at smaller radius.

There are probably other factors I have missed.

I have tried to drop on the outer, and lift on the inner, and vice versa to see if I can get an increase in energy, I have also let the mass change pulleys half way so the mass's inertia from standstill is less of an effect. Fletcher gives good advice. If you think you have something interesting or abnormal, post good pics and a vid for accurate feedback.

Cheers (wff)T

[Fletcher]

Pequaide .. Ruggerrodk asked the question twice because he got no response the first time - I, no doubt like WFT, expected you to answer him but you chose not to - I answered him second time around because he's asking for help & I pointed him in the right direction so he can draw his own conclusions about what the experiment shows & once he understands them what to look for to analyse them correctly in future.

Perhaps you could ask Ruggerrodk to explain his findings to you.

N.B. jim_mich only agreed with you in that should a mass move closer or further away from a CoR [at constant RPM] then as the radius changes so the speed & Cf's go up & down linearly whilst the energy goes up & down by the square - clearly a mass moving outwards loses energy by the square of the displacement & energy must be inputted against Cf's to move it to a closer radius, where energy increases by the square - equally clear is that the minimum energy input required in the second scenario must come from an integral of force times displacement - WFFT also agreed with jim_mich's statement of this fundamental of wheel dynamics i.e. a force is required to shift masses radially inwards against Cf's & this will increase wheel energy, conversely a mass moving outwards with Cf's requires no input of energy [ f x d ] & loses energy proportional to the radius increase - this energy is lost to inelastic collisions, heat, sound etc.

If there were any substance to your I = mr theory then Conservation of Energy would not hold & you would be able to demonstrate it with an experiment & energy budget.

N.B. physics is built on first principles & they show symmetry in all the cases you have ever quoted.

[pequaide]

You can see why people don't try anything; you can post experiments from others that confirm your theories and WFFT still attach and degrade you. They are winning; I will make no bones about it ; they are winning. But the losers are everybody else on the site. The others know they can't try anything or say anything; the best thing they can do and most have; is leave. And that too is what they want; and then there is no one with energy producing experiments. And then they control their own little dark world.

[Fletcher]

Ruggerodk .. it's quite easy to do an experiment to prove the I = mr^2 or I = mr theory - use the experiment you are discussing if you wish.

Wrap the nylon around the shaft but don't tie it to the shaft - the drive weight falls accelerating the system(s) - when it has fallen a chosen height [hits a stop] it will have lost X Pe - the drive weight will have different Y Ke - and the flywheel with masses attached at different radii for comparison will have Z Ke - Y + Z should equal X energy but this is what you are going to find out, or not.

N.B. X Pe will be the same in both experiments - Y & Z Ke's will be different for both experiments & you will see that in the time it takes for the drive weight to fall.

You don't need to know what Y & Z are for each experiment - all you need to know is that you now have a free-wheeling flywheel - and a flywheel that has Ke also stores potential for release.

All you have to do is think up a way to use that energy to lift upwards a LOAD (that could be the same mass as your driver) - this is called a Lift Test & it measures Pe gained in the Load - if the load can lift higher than the driver was dropped (for the same flywheel rpm) then you have proven that the relationship is I = mr as pequaide postulates & created GPE in the process, & this is potential energy available to do external work, & proves that CoE Law is violated, in one experiment.

If you want to know how to design a lift test from a moving flywheel with shaft then ask the community as I'm sure there will be some good ideas - VProject1 uses a nice engagement lever in his Bessler Wheel video, but he would know a good method.

Some would just say why can't you hang a drive weight from a circular chain to drive up the flywheel & then when the driver is past the lowest point it climbs again - if it gets higher you have created energy - oh yeah, it has been said before.

ETA: I probably need to add this to tie it back to your time taken query - if you gear up the drive weight to the flywheel so the flywheel gains max rpm & Ke whilst the drive weight drops very very slowly with little as possible Ke then it should be a cinch to lift the load higher than the driver start height - case proven that I = mr, all momentum can be transferred, & CoE is false.

[pequaide]

http://www.bbc.co.uk/schools/gcsebitesi ... rev1.shtml

They actually call it the velocity ratio, I am sure they know the difference between distance and velocity. One kilogram moving five meters per second is being treated as the same as 5 kilograms moving one meter per second. They do not state that it takes 5 time as much force to move 1 kilogram at 5 meters per sec, because of course it does not. Are these guys frauds too?

This is how levers work; it is the velocity ratio. You can either have 5 kilograms moving 1 m/sec or 1 kg moving 5 m/sec. It is the velocity ratio.

[Wubbly]

http://lecturedemo.ph.unimelb.edu.au/Me ... of-Inertia

I measured the system again and I got about a 5.7 radius relationship.

I also noticed that the short radius begins to discontinue acceleration at about 110 frames in. So I compared the angular changes for both at about 100 frames in. This stall is probably caused by air resistance.

At around 100 frames the short radius is covering 360° in 11 frames; and the long radius takes about 65 frames. 65/11 = 5.9

And 5.7 squared is 32.49.

So is 5.9 closer to 5.7 or 32.49

... you can post experiments from others that confirm your theories and WFFT still attach and degrade you. They are winning;...

http://lecturedemo.ph.unimelb.edu.au/Me ... of-Inertia

Is this the experiment, from others, that supposedly confirms your theory, pequaide?

The I=mr hypothesis predicts the acceleration (not to be confused with angular acceleration) will be the same for both cases.

Since the distance is 5.7 times as long, and using t=SQRT(2d/a) to calculate the time, the I=mr hypothesis predicts the time will be longer by a factor of the square root of 5.7; which is 2.4 times longer.

Your theory predicts 2.4 times longer and your analysis of the experiment comes in at 5.9 times longer.

How does this experiment confirm your theory, pequaide? It's not even in the ballpark.

[pequaide]

I thought the best way to learn how and where to blog is to look at other blogs. I found that there are others that have sims that show centrifugal transfer will work. The blogs may be a better place to be.

[pequaide]

Mark: I agree with your frustrations; I think the only thing to do is ignore them. To refocus; I repeated an old experiments. It was the 18.8 inch diameter plywood wheel. The shaft is .75 inches; the extra diameter comes from the rope diameter.

I placed one 20.45 kg mass on each side of the shaft and accelerated it with an extra 3.4 kg (7.5 lbs) at the shaft. I used a stop watch to record three runs of a single rotation each. The three runs for a single rotation were 3.10 sec; 3.15sec; and 3.07 sec.

I then placed one .898 kg mass on each side of the outer diameter of the wheel and accelerated it with an extra 3.4 kg (7.5 lbs) at the shaft. I used a stop watch to record three runs of a single rotation each. The three runs for a single rotation were 2.78 sec; 2.88 sec; and 2.85 sec.

Even though the data is rough and too fast it clearly shows that the 3.4 kg of accelerating mass is treating the 20.45 kg at .414 inches the same as .900 kg at 9.4 inches.

And because the .900 kg is moving a whole lot faster (22.7 times as fast) it has a whole lot more energy.

The total mass accelerated was 40.9 kg at the shaft and 1.8 kg at the outer diameter. Every thing else was the same and was accelerated to the same speed; the only difference is these two mass quantities. The WFFT theory would not move this fast until the mass on the outside diameter is only .040 kg.

This has to do with the other busy post where Jim has those arrows in his back. I predicted 2.18 sec or so; and they are predicting around 21 sec. I notice they don't even mention my prediction. They apparently all live in sim land.

With a mass-less wheel and ideal conditions the 2.18 seconds is correct.

This is the Old post; The two wheels in the double Atwood’s could be looked at as two lever arms. If the shaft has a working diameter of .828 inch and the large wheel has a diameter of 18.8 inches then the lever advantage is 22.7 to one. We can suspend 20.45 kilogram from one side of the shaft and 20.45 kilograms from the other side of the shaft. We can then accelerate the two suspended masses with an additional 3.4 kg suspended from one side. We will get an F = ma relationship for the acceleration of the 44.3 (20.45 + 20.45 + 3.4) kilograms being accelerated by 3.4 kg (3.4 kg * 9.81 N/kg / 44.3 kg =.753 m/sec/sec). How close we get to a perfect F = ma will be determined by how good our bearings are and how small we can keep the rotational inertial of the wheel. We could put the motion of the wheel in as part of the F = ma equation but how we apply this equation to the motion is what is being discussed. So for now it would probably be best to ignore the inertia of the wheel.

The 3.4 kilogram accelerating the 44.3 kilograms will give us an acceleration of about .753 m/sec/sec. The 3.4 kilograms would have to drop .664 meter to give the system a velocity of 1 m/sec. That would give you ½ * 44.3 kg *1 m/sec * 1 m/sec = 22.15 joules of energy. This would be equal to the input energy of 3.4 kg * 9.81 N/kg * .664 meters = 22.14 joules.

What would the acceleration be if we replaced one of the 20.45 kilogram masses suspended from the shaft with a balancing mass of .9 kilograms suspended from the circumference of the wheel?

Would the accelerating mass of 3.4 kilograms back down the acceleration because it would be producing too much energy?

Would the 3.4 kilograms detect the difference in torque between 20.45 kilograms at .828 inch as opposed to .9 kilograms at 18.8 inches? Is there a torque difference between 20.45 kilograms at .828 inch as opposed to .9 kilograms at 18.8 inches?

My data shows that the 3.4 kilogram drive mass will rotationally accelerate 20.45 kg on one side and .9 kg on the other side at the same rate as 20.45 kg on both sides. This would be with 20.45 kg at the .828 inch shaft and .9 kg at the 18.8 inch circumference.

F = ma is satisfied with this because .9 kg * 22.7 m/sec = 20.43 units of momentum and 20.45 kg * 1 m/sec = 20.45 units of momentum. At any rotational velocity the suspended mass at the circumference is moving 22.7 times faster than the suspended mass off of the shaft. I picked 1 m/sec and 22.7 m/sec.

My data also shows that two .9 kg masses at the circumference accelerate the same as two 20.45 kg at the shaft.

But: ½ * 1.8kg *22.7 m/sec * 22.7 m/sec = 463.76 joules: and we only put in 22.14 joules. After a drop of .664 m the shaft is moving 1 m/sec and the circumference is moving 22.7 m/sec.

Here is a thought: In the double Atwood's, with 20.54 kilograms on the shaft and .8964 kilograms balanced on the other side at the circumference, the energy remains the same after it moved. If the 20.54 kilograms goes down one centimeter then the .8964 kilograms goes up 22.9 cm. The 20.54 kg * 9.81 N/kg * .01 m = 2.015 joules of energy. The .8964 kg * 9.81 N/kg * .229 m = 2.014 joules.

But while they are moving the energy is not the same. If the 20.54 kilograms is moving .01 m/sec then the .8964 kilograms is moving .229 m/sec; for .001027 joules and .0235 joules. It takes 22.9 times as much energy for the .8964 kg to have the same energy change. In other words the Law of Conservation of Energy disproves itself. How can you put in different quantities of energy and get the same energy out? What kind of conservation law is that?

Suggestions on other places to post would be welcome; nothing will be read from WFFT.

[pequaide]

Pictures of the plywood wheel can be found by searching 'plywood wheel', May 10, 2012: great job Scott.

[Wubbly]

http://www.besslerwheel.com/forum/viewt ... 9377#99377

Plywood wheel with a 0.75" diameter shaft = 0.375" radius shaft = (0.009525 meter radius).
Hang 20.45 kg off each end of the shaft. They balance.
Hang an additional 7.5 pounds (3.4 kg) on one side of the 0.009525 meter radius.
Let the wheel rotate 180 degrees.
It takes about 2 seconds.
The distance travelled by the masses is half the circumference = 1/2 * 2 PI r = 0.5 * 2 * PI * 0.009525 meters = 0.03 meters.
The acceleration of the system is a=2d/t² = 2 * 0.03 meters / (2 seconds)² = 0.015 m/s²
The input force of the driver mass is F=mg = 3.4 kg x 9.81 m/s² = 33.4 Newtons.
And since F=ma, the overall mass of your system is m = F/a = 33.4 N / 0.015 m/s² = 2227 Kilograms.

Your system is accelerating like it has a mass of 2227 kilograms, but you want us to focus on two 20.45 kg masses, and a 3.4 kg mass, and ignore the fact that 98% of your system mass in unaccounted for.

Where is the missing mass?

So for now it would probably be best to ignore the inertia of the wheel.

Oh, there it is. You want us to ignore it, ignore the fact that you are accelerating your wheel from a 0.009525 meter radius, and pretend the rotational mass of the plywood and the tight driver radius does not affect the acceleration of the masses you hang off it.

pequaide, there's a bus in your experiment and you want us to focus on the bowling ball and the pea.

http://www.besslerwheel.com/forum/viewt ... pea#101730

[pequaide]

It must be a sad life to have nothing more to do with your mind than to stalk people on the internet.

[Wubbly]

Not as sad as believing you can create energy by dropping 3.4kg a total of 3 centimeters, and missing 98% of your experiment.

Not as sad as not being able to learn a single thing in the last 5 years.

[Dunesbury]

It is good exercise lifting the weight for the experiments so not a total loss.

[pequaide]

A Law is a Law all of the time; A Law can’t pick and choose what days it feels like being right.

All the potential energy of the 3.4 kilogram is either held in the motion or is lost to the bearings (heat?) and air resistance.  So you have 100% in and 100% out. The motion of the 40.9 kilograms, and the motion of the 3.4 kilograms, and the motion of the wheel; plus the lost energy in the bearing and to air: equals 100% of the original energy.

I calculate about 2.22 joules of input energy. The 40.9 kilograms has a final velocity of .043 m/sec for an energy of .037 joules. So the other energy remaining in the system or lost by the system is 2.138 joules.

The ‘other energy remaining or lost in the system’ is the same for the 1.8 (.900 * 2) kilogram run; 2.138 joules. It is the same 2.138 joules because it is the same bearing and the same dropped mass is moving at the same speed; the wheel is moving at the same speed; and air resistance should be about the same.

The 1.8 kilograms (at the circumference) has a final velocity of over one meter per second, for an energy of almost one joule.  1 joule plus 2.138 joules is 3.138 joules; which is 146% of the original energy.

If F = ma is taken as a conserved Law then no change arises; 40.9 kilograms moving .043 m/sec is roughly equal to the momentum of 1.8 kilograms moving 1.04 m/sec. And F = ma passes the test of actually being right.

According to the other theory the rotation should not be at these speeds until the perimeter mass is down to 40 grams on each side. So lets do 100 grams on each side and it will be going too fast, so the fact that the wheel has mass is no excuse the system is moving too fast. The wheel inertia does not make it go faster. The only way to make it go faster is to remove inertia; and obviously 100 grams on the perimeter is less inertia than 20.45 kilograms at the shaft.

I am convinced that there is nothing you can say; or experiments that you can show that will have any effect upon the four that control this site.  Moving on is an accepted idea; even though I very much appreciate to format of this site.

[pequaide]

If you use the key words “angular momentum conservation videos� you will find a place (Khan Academy) that states that linear momentum can be increased without the application of outside force.   He states that a rotation of 10 radians /sec changes to 40 radians /second in a half size circle. This is the only rotation speed that would conserve angular momentum. And he is so certain that angular momentum is conserved that he does not stop to think that in order for the mass to rotate four times as fast it will have to have a doubling of arc (or linear) velocity.
 
He states that this doubling of arc velocity is done without the application of outside torque; or outside force. We experimentally know that arc velocity and linear velocity are equal; and the doubling of arc velocity is a quadrupling of energy.

But I have a question: why doesn’t the instructor reduce the radius to one tenth? Then the linear velocity would, according to him, increase to ten times as fast.  Or why not reduce the radius to one hundredth; then we could use the 100 times linear velocity to hunt big game.

This is another case of the mr² theory; and it simply does not work. The only (none) experiments that are given is the ice skater. They don’t evaluate the mass or distribution of mass. They don’t measure actual speed or the radii. And they ignore the fact that the mass radius is reduced by using an outside force (muscle). 

Real experiments like Galileo’s pendulum or a double ballistic pendulum or the aluminum rod are ignored and ridiculed.

But this is the chink in their armor that you are looking for. Angular momentum conservation does not apply to objects in the lab; where there is no, deep space like, gravitational acceleration for the mass.

Almost no one will experiment in this area because of a devout belief in the Law of Conservation of Angular Momentum.

A spinning wheel will violently throw smaller objects; and it is mv conservation not angular momentum conservation.