Lets use 1 m/sec as the initial arc velocity of the combines mass. Then each reader should support a number for the velocity of the spheres when they have all the motion. And let the reader support with experiments and Laws the reasons for supporting 9 m/sec or 3 m/sec.
I pick 9 m/sec because it is supported by The Law of Conservation of Momentum; ballistic pendulum experiments; and the physical measurement of the sphere's velocity.
What Law do you use Me?; seeing that there is no law of conservation of kinetic energy.
energy producing experiments
Moderator: scott
The reader should be able to verify the measurement, it's not about choice or picking a formula which is the most convenient.
When you find a relation in some spread of measurements, then the correct formula should emerge or match an existing one.
I could probably write a nice poem explaining which kind of "Law" to use or which "Law" would be my favorite,
But such isn't relevant at all and without any meaning: it's not a popularity contest... it's science (or it should be).
How do you measure a millimeter in the Z-plane (towards the horizon) from a perspective distorted picture of a perspective distorted recording made from the wrong point of view??
I think I detected some button which seems to indicate some "save frame"-functionality?: such would be of much help to start with.
When you find a relation in some spread of measurements, then the correct formula should emerge or match an existing one.
I could probably write a nice poem explaining which kind of "Law" to use or which "Law" would be my favorite,
But such isn't relevant at all and without any meaning: it's not a popularity contest... it's science (or it should be).
How do you measure a millimeter in the Z-plane (towards the horizon) from a perspective distorted picture of a perspective distorted recording made from the wrong point of view??
I think I detected some button which seems to indicate some "save frame"-functionality?: such would be of much help to start with.
re: energy producing experiments
How must I know?
M=Mass Cylinder
R=Radius of the cylinder
m=mass of a single sphere
r=length of the tether
f=linear speed-up factor
f=Sqrt( (M+2m)/(2m)) = (R+r)/R
The situation is probably where I say 3 and you claim 9: I'm happy with both outcomes.
How will you proof it, as your latest picture doesn't show anything at all.
- What's the weight of the cylinder
- What's the inner diameter if the cylinder
- What's the outer diameter of the cylinder
- What's the weight of a single sphere
- What's the diameter of of a single sphere
- What's the length of the tether (measured from where to where)
- What's the initial rotational speed
- Does the cylinder keep rotating when the tether is not deployed?
M=Mass Cylinder
R=Radius of the cylinder
m=mass of a single sphere
r=length of the tether
f=linear speed-up factor
f=Sqrt( (M+2m)/(2m)) = (R+r)/R
The situation is probably where I say 3 and you claim 9: I'm happy with both outcomes.
How will you proof it, as your latest picture doesn't show anything at all.
Marchello E.
-- May the force lift you up. In case it doesn't, try something else.---
-- May the force lift you up. In case it doesn't, try something else.---
re: energy producing experiments
All of the questions have been answered so you are obviously not paying attention.
I used tape on the monitor and marked it with ink. There are five frames between 165 and 170, just like there are five frames between 225 and 230, looks like Newton wins; momentum is conserved.
You need nine times the initial energy at 197 to return to and achieve the final energy at 227.
When people reference the law of conservation of energy, the second thing they sight is why it does not work experimentally. “Oh’ well we lose heat we cannot find�.
Well how does that heat stuff work out for you when the motion of the combined mass is returned in this experiment?
It is obvious that nothing is being lost.
I used tape on the monitor and marked it with ink. There are five frames between 165 and 170, just like there are five frames between 225 and 230, looks like Newton wins; momentum is conserved.
You need nine times the initial energy at 197 to return to and achieve the final energy at 227.
When people reference the law of conservation of energy, the second thing they sight is why it does not work experimentally. “Oh’ well we lose heat we cannot find�.
Well how does that heat stuff work out for you when the motion of the combined mass is returned in this experiment?
It is obvious that nothing is being lost.
re: energy producing experiments
That 1:9 was for some cart wheel, this is a pvc cylinder - how must I know what experiment you show here?
If you want verification, you should provide it for each and every situation as if there was no before. I have never seen a post where you list those parameters all in a single post, or otherwise scattered between text.
You used tape: why?
What does "five frames" at 240 fps mean besides 20.833... milliseconds?
You didn't measure energy, you conclude it from a formula.
I think your use of formula is wrong - just as you think my formula usage is wrong.
Anyway your "joules" are the ones to be verified, luckily we could measure speed to get an indication.
Show the pictures where speed can be verified (perhaps top-view sequences are too much to ask), and I will gladly go sweating over my own formulas to see how I could have been so very wrong.
When I include the estimated relative sizes I can stretch things to a factor of 3.5 maybe 3.6, so seeing things at about 9 or down to maybe as low as 7 would be relatively easy to spot but unexpected from my theoretical view.
As far as I can see there should be no noticeable heat-up processes in your spheres, nor should there be a huge collision with your cylinder.
A full return is exactly what's predicted according to current physics.
But that's explained before, obviously you're not paying attention either.
If you want verification, you should provide it for each and every situation as if there was no before. I have never seen a post where you list those parameters all in a single post, or otherwise scattered between text.
You used tape: why?
What does "five frames" at 240 fps mean besides 20.833... milliseconds?
You didn't measure energy, you conclude it from a formula.
I think your use of formula is wrong - just as you think my formula usage is wrong.
Anyway your "joules" are the ones to be verified, luckily we could measure speed to get an indication.
Show the pictures where speed can be verified (perhaps top-view sequences are too much to ask), and I will gladly go sweating over my own formulas to see how I could have been so very wrong.
When I include the estimated relative sizes I can stretch things to a factor of 3.5 maybe 3.6, so seeing things at about 9 or down to maybe as low as 7 would be relatively easy to spot but unexpected from my theoretical view.
As far as I can see there should be no noticeable heat-up processes in your spheres, nor should there be a huge collision with your cylinder.
A full return is exactly what's predicted according to current physics.
But that's explained before, obviously you're not paying attention either.
Marchello E.
-- May the force lift you up. In case it doesn't, try something else.---
-- May the force lift you up. In case it doesn't, try something else.---
re: energy producing experiments
No it is not predicted unless NASA is not part of the accepted world of physics.
NASA predicted 20 m/sec for the 3 kilograms not 400 m/sec.
It will take 400 m/sec to return the satellite back to it original rate of rotation.
What's the weight of the cylinder 2432 g
What's the inner diameter if the cylinder 2.81 inches
What's the outer diameter of the cylinder 3.5 inches
What's the weight of a single sphere 152 g
What's the diameter of a single sphere 1 inch
What's the length of the tether (measured from where to where) two r (3.5 inches) side wall to center of the sphere
What's the initial rotational speed about 3.5 rps
Does the cylinder keep rotating when the tether is not deployed? Of course
NASA predicted 20 m/sec for the 3 kilograms not 400 m/sec.
It will take 400 m/sec to return the satellite back to it original rate of rotation.
What's the weight of the cylinder 2432 g
What's the inner diameter if the cylinder 2.81 inches
What's the outer diameter of the cylinder 3.5 inches
What's the weight of a single sphere 152 g
What's the diameter of a single sphere 1 inch
What's the length of the tether (measured from where to where) two r (3.5 inches) side wall to center of the sphere
What's the initial rotational speed about 3.5 rps
Does the cylinder keep rotating when the tether is not deployed? Of course
re: energy producing experiments
@ ME .. these questions have been asked multiple times, even by you recently.
1. How much quantitative energy is expended giving the cylinder and spheres experiment motion (to an rpm) ?
2. How does that initiation energy expenditure (1. above) compare to various downstream positions of the cylinder and spheres i.e. KE's after motion is initiated, stable rpm, and the spheres are then released ?
3. Is there an excess of usable Output energy (capacity to do work) above Input energy of 1. above to indicate possible OU, even after factoring for possible friction losses ?
N.B. this assumes that any system loss of GPE is negligible or has been factored into the results. And that the experiment is free from human error and bias.
If those questions were ever answered then it would be interesting to compare it to your formula predictions.
1. How much quantitative energy is expended giving the cylinder and spheres experiment motion (to an rpm) ?
2. How does that initiation energy expenditure (1. above) compare to various downstream positions of the cylinder and spheres i.e. KE's after motion is initiated, stable rpm, and the spheres are then released ?
3. Is there an excess of usable Output energy (capacity to do work) above Input energy of 1. above to indicate possible OU, even after factoring for possible friction losses ?
N.B. this assumes that any system loss of GPE is negligible or has been factored into the results. And that the experiment is free from human error and bias.
If those questions were ever answered then it would be interesting to compare it to your formula predictions.
re: energy producing experiments
This is from an Aug 20 post; I don't know why this does not make you a prevaricator.
The rotational energy at 167 and 227 is ½ * 2.736 kg * .96m/sec * .96m /sec (4 mm/frame * 240 frames/sec) = 1.26 joules.
That means that the energy at 197 is 9 * 1.26 joules or 11.346 joules.
I suppose you thing it is okay to not even read the person's post but then complain about them.
Flet; you have been asked many times not to post on this thread.
Flet: please do not post on this thread.
The rotational energy at 167 and 227 is ½ * 2.736 kg * .96m/sec * .96m /sec (4 mm/frame * 240 frames/sec) = 1.26 joules.
That means that the energy at 197 is 9 * 1.26 joules or 11.346 joules.
I suppose you thing it is okay to not even read the person's post but then complain about them.
Flet; you have been asked many times not to post on this thread.
Flet: please do not post on this thread.
re: energy producing experiments
What are you upset about Peq? Fletchers questions and comments are calm, logical and relevant.
You have not posted publicly your video for scrutiny by those that might be interested. I think this makes your job of convincing people much more difficult. Also take a step back, and read your posts from an outsiders perspective. To me, your posts often are difficult to interpret.
You have not posted publicly your video for scrutiny by those that might be interested. I think this makes your job of convincing people much more difficult. Also take a step back, and read your posts from an outsiders perspective. To me, your posts often are difficult to interpret.
re: energy producing experiments
The original energy is not only in print (1.26 J) but it is in photograph; in 1:07:165 the mark on the tape on the monitor is on one side of a 20 mm square. At 170 the photograph shows that 5/240th of a second later it is on the other side. It has moved 20 mm in 5/240th of a second. 20 mm / X = 5/240 is .96 m/sec. That .96 m/sec is then placed in the kinetic energy formula which is ½ * 2.736 kg * .96 m/sec * .96 m/sec = 1.26 joules
This 1.26 joules was attributed to the locations of 1:07:167 and 1:07:227 ; because the same quantity of motion is measured at 227 (also in photos).
For a one ninth mass to give a larger mass 1.26 joules of energy; it will have to have nine times that much energy. From ballistic pendulums we know that the .304 kg spheres will have to have 11.35 joules of energy. And that 11.35 J was attributed to 1:07:197.
That is an energy increase to 900% of the original energy.
Overheads are good but the motion is blurred because of the high speed. The slow motions are well accepted but I need direct contact or addresses; to send the flash drives.
This 1.26 joules was attributed to the locations of 1:07:167 and 1:07:227 ; because the same quantity of motion is measured at 227 (also in photos).
For a one ninth mass to give a larger mass 1.26 joules of energy; it will have to have nine times that much energy. From ballistic pendulums we know that the .304 kg spheres will have to have 11.35 joules of energy. And that 11.35 J was attributed to 1:07:197.
That is an energy increase to 900% of the original energy.
Overheads are good but the motion is blurred because of the high speed. The slow motions are well accepted but I need direct contact or addresses; to send the flash drives.
re: energy producing experiments
I have my own thoughts on your latest accusation...
"This is from an Aug 20 post"... ok true.
I thought your 1:9 argument only counted for the cart-wheel; Plus in that same post you assumed energy from an another unfounded assumption with a measurement almost impossible to make, hence I discarded that whole post.
Anyway, you don't read my posts either.
One thing: how do you measure your "millimeters" in the z-plane?
If you can drag'n-drop things to some flash-drive, you could drag'n-drop it to that online thing I showed.
Until then we can enjoy an animation or two.
"This is from an Aug 20 post"... ok true.
I thought your 1:9 argument only counted for the cart-wheel; Plus in that same post you assumed energy from an another unfounded assumption with a measurement almost impossible to make, hence I discarded that whole post.
Anyway, you don't read my posts either.
One thing: how do you measure your "millimeters" in the z-plane?
If you can drag'n-drop things to some flash-drive, you could drag'n-drop it to that online thing I showed.
Until then we can enjoy an animation or two.
- Attachments
-
-
Marchello E.
-- May the force lift you up. In case it doesn't, try something else.---
-- May the force lift you up. In case it doesn't, try something else.---
re: energy producing experiments
I like your overlapping pictures.
It is interesting to note that the spheres are maintaining their position in relationship to the cylinder; they may be moving forward or closing in on the cylinder. But they are primarily doing what I call; flying in formation. This actually surprised me. While I was still primarily releasing, and before the high speed camera; I would have guessed that they would come back around and click the side of the cylinder if they were not released. But they do not tough the cylinder.
Because of the sphere position at a slightly larger radius than when they were released; they will be moving slightly faster than when released. This would give less motion to be contained in the cylinder; but the difference is still not noticeable. The speed is obtained from the number of frames needed to cross the 20 mm square on the surface of the cylinder; and the spheres would be moving above that speed.
If the sphere motion (when they have all the motion) conserved the original energy then only a third of the necessary motion would be available to cause the five frame crossing. The five frame crossing would turn into a 15 frame crossing.
I am curios what measurement was a “measurement almost impossible to make�?
How is this for a momentum conservation corollary? When a mass of 1/n shares its motion with a combined mass of n; the quantity of motion energy is reduced to 1/n th.
Plugging in numbers we would have: When 1 kilogram moving nine meters per second shares its motion with 8 other kilograms; the new velocity would be 1 m/sec. The original energy was 40.5 J and the new energy is 4.5 J. 4.5 / 40.5 = 1/9
It is interesting to note that the spheres are maintaining their position in relationship to the cylinder; they may be moving forward or closing in on the cylinder. But they are primarily doing what I call; flying in formation. This actually surprised me. While I was still primarily releasing, and before the high speed camera; I would have guessed that they would come back around and click the side of the cylinder if they were not released. But they do not tough the cylinder.
Because of the sphere position at a slightly larger radius than when they were released; they will be moving slightly faster than when released. This would give less motion to be contained in the cylinder; but the difference is still not noticeable. The speed is obtained from the number of frames needed to cross the 20 mm square on the surface of the cylinder; and the spheres would be moving above that speed.
If the sphere motion (when they have all the motion) conserved the original energy then only a third of the necessary motion would be available to cause the five frame crossing. The five frame crossing would turn into a 15 frame crossing.
I am curios what measurement was a “measurement almost impossible to make�?
How is this for a momentum conservation corollary? When a mass of 1/n shares its motion with a combined mass of n; the quantity of motion energy is reduced to 1/n th.
Plugging in numbers we would have: When 1 kilogram moving nine meters per second shares its motion with 8 other kilograms; the new velocity would be 1 m/sec. The original energy was 40.5 J and the new energy is 4.5 J. 4.5 / 40.5 = 1/9
Glad you liked it..
From a rough estimation they indeed seem to follow formation, so it looks like it goes around 1m/s.
*Because of an ellipse for a single frame, you don't film heads-on and slightly from above: this means the path of your sphere also show up as some ellipse.
*This ellipse changes in shape between 5 frames: this means there's a significant perspective distortion between those frames.
A single millimeter is not the same distance when the sphere peeps just from behind the cylinder compared to the distance when it flies in front of the sphere.
Because of this perspective there's at least a parallax issue when using the cylinder as reference, even for a single frame and even more between frames.
It would be somewhat do-able when you're sure the spheres followed a circular path. And they don't as that's the whole purpose of your experiment.
Then there's an indication of barrel-distortion from your camera lens as your 230-sphere is much bigger than your 225-sphere, meaning: you can't extrapolate in straight lines in the Z-plane.
The "measurement" you make is based on an observation which could be an optical illusion: thus, how do you know (measure) it moves in formation.
Hence: how do you measure a millimeter in 3D-space from a 2D-picture with all kinds of uncertainties.
Attached is a possible 3D-situation sketch where I tried to match the ellipse. The ellipse seems good, but also indicates an additional height-drop when presented this way.
Either this or some distortion, but both possibilities have their own issues while assuming a constant radius.
From a rough estimation they indeed seem to follow formation, so it looks like it goes around 1m/s.
As you hopefully noticed the top ellipse-shape of the cylinder changes, this indicates some issues:I am curios what measurement was a “measurement almost impossible to make�?
*Because of an ellipse for a single frame, you don't film heads-on and slightly from above: this means the path of your sphere also show up as some ellipse.
*This ellipse changes in shape between 5 frames: this means there's a significant perspective distortion between those frames.
A single millimeter is not the same distance when the sphere peeps just from behind the cylinder compared to the distance when it flies in front of the sphere.
Because of this perspective there's at least a parallax issue when using the cylinder as reference, even for a single frame and even more between frames.
It would be somewhat do-able when you're sure the spheres followed a circular path. And they don't as that's the whole purpose of your experiment.
Then there's an indication of barrel-distortion from your camera lens as your 230-sphere is much bigger than your 225-sphere, meaning: you can't extrapolate in straight lines in the Z-plane.
The "measurement" you make is based on an observation which could be an optical illusion: thus, how do you know (measure) it moves in formation.
Hence: how do you measure a millimeter in 3D-space from a 2D-picture with all kinds of uncertainties.
Attached is a possible 3D-situation sketch where I tried to match the ellipse. The ellipse seems good, but also indicates an additional height-drop when presented this way.
Either this or some distortion, but both possibilities have their own issues while assuming a constant radius.
- Attachments
-
Marchello E.
-- May the force lift you up. In case it doesn't, try something else.---
-- May the force lift you up. In case it doesn't, try something else.---