Hippy! :DJim_Mich wrote:"Make Love, Not War."
Atwoods Analysis
Moderator: scott
re: Atwoods Analysis
re: Atwoods Analysis
With a fast old joke you jet a dark cloud and leave the core of the question fully in open!
FWIW, very evident a low rate of love/year.
Best!
Murilo
FWIW, very evident a low rate of love/year.
Best!
Murilo
re: Atwoods Analysis
Broli; I now have a 40 kilogram rim that is mounted on two pillow blocks. It has a diameter of a little over one meter. It spins very easily. If I accelerate it by hand to one meter per second, and the Law of Conservation of Momentum is true, it should be able to throw a one kilogram object 80 meters up. If on the other hand the Law of Conservation of Energy is true it should only throw the one kilogram two meters up.
A three year old could throw a ball two meters and it would scare no one, but I know better, I am staying away from the thing. I have seen the violent velocities in smaller wheels.
I will now engineer a few mechanical releases so I won’t be any where near the spinning wheel when it flings out the one kilogram missile.
Jim says he won’t experiment “They should do it themselves�. And “I (Jim) am helping them by doing nothing�. Well let me be the first to thank you for all your help Jim; Thanks.
Do you have a list of those other sites Broli?
A three year old could throw a ball two meters and it would scare no one, but I know better, I am staying away from the thing. I have seen the violent velocities in smaller wheels.
I will now engineer a few mechanical releases so I won’t be any where near the spinning wheel when it flings out the one kilogram missile.
Jim says he won’t experiment “They should do it themselves�. And “I (Jim) am helping them by doing nothing�. Well let me be the first to thank you for all your help Jim; Thanks.
Do you have a list of those other sites Broli?
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FunWithGravity2
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re: Atwoods Analysis
Pequiade,
I have followed your thread and love what you are doing, I have been curious about a potential "proof" that i wanted to put forward but since i don't get much response i assumed it would go by the wayside so i did not bother. But since you seem to have a heavy test rig and are more than knowledgeable i wonder if you would indulge.
if your 40 kilo wheel is spun up to 60 rpm and a weigth of 1 kilo is drop onto the wheel as it spins at the outer rim at 9 oclock. And its dropped from the same height as the top/12 oclock we all know this weigh will be propelled away just like in a pitching/tennis machine. But what i am curious about is whether the energy/momentum of the flywheel is reduced by the impact in direct relation to the amount of energy gained by the ball shooting off into the atmosphere.
All the flywheels that i use have been made to be very light, i could add mass but to be quite honest i have absolutely no skill to figure out how to measure this type of thing. If you are persuing this course then i look forward to your results. If not then if you have any chance to play around with it then i would love to hear your thoughts.
Keep up your excellent work.
Dave
FWIW , if you could accelerate the wheel with just an OB weight that drops of at the bottom then the experiment might be a LITTLE more determinate than accelerating the wheel by hand. I have a setup where i have an old sewing machine motor rigged to a setup so as to rotate the flywheel to test CF/CP resistance and this works very well for getting constant speed if you want to test constant speed as opposed to energy input.
I have followed your thread and love what you are doing, I have been curious about a potential "proof" that i wanted to put forward but since i don't get much response i assumed it would go by the wayside so i did not bother. But since you seem to have a heavy test rig and are more than knowledgeable i wonder if you would indulge.
if your 40 kilo wheel is spun up to 60 rpm and a weigth of 1 kilo is drop onto the wheel as it spins at the outer rim at 9 oclock. And its dropped from the same height as the top/12 oclock we all know this weigh will be propelled away just like in a pitching/tennis machine. But what i am curious about is whether the energy/momentum of the flywheel is reduced by the impact in direct relation to the amount of energy gained by the ball shooting off into the atmosphere.
All the flywheels that i use have been made to be very light, i could add mass but to be quite honest i have absolutely no skill to figure out how to measure this type of thing. If you are persuing this course then i look forward to your results. If not then if you have any chance to play around with it then i would love to hear your thoughts.
Keep up your excellent work.
Dave
FWIW , if you could accelerate the wheel with just an OB weight that drops of at the bottom then the experiment might be a LITTLE more determinate than accelerating the wheel by hand. I have a setup where i have an old sewing machine motor rigged to a setup so as to rotate the flywheel to test CF/CP resistance and this works very well for getting constant speed if you want to test constant speed as opposed to energy input.
re: Atwoods Analysis
Per Pequaide's specs... A one meter diameter 40 kilogram rim moving at one meter per second will be moving at about 19 RPM. A weight fixed to the rim and let loose will fly upward about 0.05 meter. A weight flung in an arc outward by way of a string wrapped around the rim should accelerate up to about 40 m/s while slowing down the wheel to almost a stop. (There is a bunch of fudging of the actual values going on here which I can explain if anyone really wants a bunch of geeky formulas involving radius of gyration and such.) This should send the weight up to a height of about 80 meters (about 262 feet). Of course getting the release time just right for the weight to go straight up will be tricky.
Of course the hard part is reversing your process. A weight falling from 80 meters should be able to spin the rim back up to about 1 meter per second, minus friction.
Pequaide, in the past I've tried to help individuals to understand their concepts but many times they do not like what I say, Murilo being a prime example. That is why I stated "They should do it themselves.", simply because many times some people will not believe the results unless they witness it themselves. That is why I say I'm helping. Do you understand how people can be helped to learn by doing it themselves?
Of course the hard part is reversing your process. A weight falling from 80 meters should be able to spin the rim back up to about 1 meter per second, minus friction.
Pequaide, in the past I've tried to help individuals to understand their concepts but many times they do not like what I say, Murilo being a prime example. That is why I stated "They should do it themselves.", simply because many times some people will not believe the results unless they witness it themselves. That is why I say I'm helping. Do you understand how people can be helped to learn by doing it themselves?
Re: re: Atwoods Analysis
Check your pm.pequaide wrote: Do you have a list of those other sites Broli?
re: Atwoods Analysis
Yes I am fudging just a little. I am only using the mass of the rim for my 40 kilogram calculations. The steel rim is 1.25 in. by 2 in. and the diameter to its center of mass (of the rim) is 41.5 inches. I think this gives use a rim mass of 41.66 kg. The six spokes are ¾ steel rods, I believe the radius of gyration of a spoke would be .5. The hub would have a mass of a kilogram or so but its speed is small and its radius of gyration would reduce its effect even more. So for the purpose of discussion I rounded this all off to 40 kilograms.
The point is that: if the momentum of the wheel is conserved you will get a throw of about 80 meter. If kinetic energy is conserved the missile would not even raise much above the wheel, not even the length of the tether.
I have done this with a smaller 18 inch wheel, and believe we the motion is violent.
Jim quote: Of course the hard part is reversing your process. A weight falling from 80 meters should be able to spin the rim back up to about 1 meter per second, minus friction.
Oh: but you don’t want to reverse the process after the missile has risen 80 meters. You want to place the one kilogram mass in a 40 kilogram Atwood’s and drop it one meter 80 times. But let’s do one thing at a time. Now we are throwing things.
FunWithGravity2: Yes any mass added to the spinning wheel would absorb some of the motion of the wheel. But the idea is to wrap it with a string and absorb all the motion of the wheel.
Wrap two strings around your small flywheels so that they counterbalance as they unwind. By having two strings unwind simultaneously the added force on the center bearing is almost zero.
Light flywheels work too, but you must also use light masses on the end of your string; an attached (total sphere mass) mass of about 1/5 the mass of the wheel works great. Of course the effect of air resistance for light masses is higher than it is for more massive objects.
As far as adding mass to the wheel: you add the same amount of mass the same distance from the axis on opposite sides of the wheel. You must also add the mass in the same plane as the mass of the wheel, or at equal and opposite distances from the plane. This might be a better why to visualize it: The center of mass of the added mass must be at the same point as the center of mass of the wheel.
It is easy to balance the mass of a vertical wheel. The heavy point of a vertical wheel will rotate to 6 o’clock. You merely add mass on the line to 12 o’clock until the wheel no longer has a tendency to rotate when 6 o’clock is placed at 3 o’clock.
It is all fun, as you say. Have fun.
The point is that: if the momentum of the wheel is conserved you will get a throw of about 80 meter. If kinetic energy is conserved the missile would not even raise much above the wheel, not even the length of the tether.
I have done this with a smaller 18 inch wheel, and believe we the motion is violent.
Jim quote: Of course the hard part is reversing your process. A weight falling from 80 meters should be able to spin the rim back up to about 1 meter per second, minus friction.
Oh: but you don’t want to reverse the process after the missile has risen 80 meters. You want to place the one kilogram mass in a 40 kilogram Atwood’s and drop it one meter 80 times. But let’s do one thing at a time. Now we are throwing things.
FunWithGravity2: Yes any mass added to the spinning wheel would absorb some of the motion of the wheel. But the idea is to wrap it with a string and absorb all the motion of the wheel.
Wrap two strings around your small flywheels so that they counterbalance as they unwind. By having two strings unwind simultaneously the added force on the center bearing is almost zero.
Light flywheels work too, but you must also use light masses on the end of your string; an attached (total sphere mass) mass of about 1/5 the mass of the wheel works great. Of course the effect of air resistance for light masses is higher than it is for more massive objects.
As far as adding mass to the wheel: you add the same amount of mass the same distance from the axis on opposite sides of the wheel. You must also add the mass in the same plane as the mass of the wheel, or at equal and opposite distances from the plane. This might be a better why to visualize it: The center of mass of the added mass must be at the same point as the center of mass of the wheel.
It is easy to balance the mass of a vertical wheel. The heavy point of a vertical wheel will rotate to 6 o’clock. You merely add mass on the line to 12 o’clock until the wheel no longer has a tendency to rotate when 6 o’clock is placed at 3 o’clock.
It is all fun, as you say. Have fun.
re: Atwoods Analysis
pequiade .. a couple of observations about your commentary - you spin the 40 kg atwoods up to 1 m/s [~19 rpm] by hand - obviously you are inputting the energy to do that - that is your starting conditions for the experiment ?!
You have designed a deployment mechanism [wound circumference rope tethers ?] to attach to your atwoods that unwinds - is it a one tether system for deployment or two opposing arms/tethers, since you mention symmetry & center bearing load minimization to dave ? - I'm imagining just a single 1 kg mass, but perhaps not ?
I believe you talk about releasing one 1 kg mass to see how high it will be flung vertically & this stopping or near stopping the flywheel.
Then you say that it doesn't matter about resetting the device up to original starting conditions of ~19 rpm at this stage.
How will dropping a 1 kg mass 1 meter 80 times achieve the desired rpm [from stopped] & harness & do extra mechanical work ?
P.S. as you have pointed out in different threads a trebuchet also doesn't just use a dropping large mass to lever a smaller mass into the air with great velocity - it uses a rope tether to extend that leverage effect - in the process it takes some of the trebuchet's momentum & turning that into velocity of the hurled mass, allowing it to go higher & further depending on trajectory.
BUT .. the counter weight driving mechanism must be reset to repeat the process & I don't see the hurled mass able to be harnessed to do that in a trebuchet, after losses.
You have designed a deployment mechanism [wound circumference rope tethers ?] to attach to your atwoods that unwinds - is it a one tether system for deployment or two opposing arms/tethers, since you mention symmetry & center bearing load minimization to dave ? - I'm imagining just a single 1 kg mass, but perhaps not ?
I believe you talk about releasing one 1 kg mass to see how high it will be flung vertically & this stopping or near stopping the flywheel.
Then you say that it doesn't matter about resetting the device up to original starting conditions of ~19 rpm at this stage.
The 1 kg mass accelerating under gravity from a relative height of 80 meters hypothetically achieved via a cart & gearing solution is surely required to get your stopped or near stopped atwoods up to 19 rpm again to fling the released mass 80 meters again for a second time & therefore replenish the energy into the flywheel that it started the experiment with ?pequaide wrote:Oh: but you don’t want to reverse the process after the missile has risen 80 meters. You want to place the one kilogram mass in a 40 kilogram Atwood’s and drop it one meter 80 times. But let’s do one thing at a time. Now we are throwing things.
How will dropping a 1 kg mass 1 meter 80 times achieve the desired rpm [from stopped] & harness & do extra mechanical work ?
P.S. as you have pointed out in different threads a trebuchet also doesn't just use a dropping large mass to lever a smaller mass into the air with great velocity - it uses a rope tether to extend that leverage effect - in the process it takes some of the trebuchet's momentum & turning that into velocity of the hurled mass, allowing it to go higher & further depending on trajectory.
BUT .. the counter weight driving mechanism must be reset to repeat the process & I don't see the hurled mass able to be harnessed to do that in a trebuchet, after losses.
Last edited by Fletcher on Sat Jan 16, 2010 10:16 pm, edited 1 time in total.
re: Atwoods Analysis
''With a fast old joke you jet a dark cloud and leave the core of the question fully in open!
FWIW, very evident a low rate of love/year.
Best!
Murilo''
( second mail )
FWIW, very evident a low rate of love/year.
Best!
Murilo''
( second mail )
re: Atwoods Analysis
A one kilogram mass need only drop 2.08 meters to accelerate a 40 kilogram Atwood’s to 1 meter per second. So you drop it 2.08 meters 40 times. This is a little under 19 rpm because the wheel diameter is a little larger than one meter
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nicbordeaux
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re: Atwoods Analysis
And each time a weight drops it flings a small one higher again ? So your theoretical runtime is what ?
If you think you have an overunity device, think again, there is no such thing. You might just possibly have an unexpectedly efficient device. In which case you will be abducted by MIB and threatened by aliens.
re: Atwoods Analysis
Well pequaide .. it will be good if you are able to put some empirical data to the theory by way of good experimental evidence - then you will see how close to your predictions things actually were, & more importantly why, if they are not close - of course that applies to the likes of me as well - every experiment is a chance to learn something.
Re: re: Atwoods Analysis
At least you pulled yourself straight. This forum on the other hand does not know the definition of empirical data and the scientific method (at least the classic one).Fletcher wrote:Well pequaide .. it will be good if you are able to put some empirical data to the theory by way of good experimental evidence - then you will see how close to your predictions things actually were, & more importantly why, if they are not close - of course that applies to the likes of me as well - every experiment is a chance to learn something.
"How do you input the energy, how does it reset, how big is the build" These type of questions are utterly useless if you haven't proved the theory. Creating energy by kilojoules should be amazing by itself but not on this forum it isn't, they want to know how to reset it...apparently the idea of an experiment creating huge amounts of energy out of thin air doesn't phase them.
I agree that there's no single person that can deliver the whole package. That is build the mechanism correctly, then setup the data gathering instruments and then process the obtained data. But that is what an open forum is supposed to be for! Do you see why I'm upset now about these fools that come here and tell me how I should do everything on my own? They have achieved nothing, most live in misery and are chasing a hopeless dream, yet they dare to speak on how to achieve success.
I have many skills but building is not one of them. I could for example gather the data, process it, do the math and present it very nicely but without a group effort I'm stuck in limbo. People here do not appreciate the value of collaboration and open development. This is why this forum has gone NOWHERE since it was made. Everyone is in it for his own good and everyone else can suck it. ME ME ME ME. Of course no one will admit that as delusion wins at the end of the day.
300 years of failure is buried under your feet people. Isn't it obvious things have to be done differently to succeed? Or will you just fuel perpetual failure until you die. Either change or perish.
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nicbordeaux
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re: Atwoods Analysis
We all choose our hobbies broli. Not making a grav wheel or anything else isn't going to kill me. I'll potter around in my grotty little shop having fun and let the genius' like you save the world. But you never know, I might beat you to it :) And you know what ? I'll save you too.300 years of failure is buried under your feet people. Isn't it obvious things have to be done differently to succeed? Or will you just fuel perpetual failure until you die. Either change or perish.
No kidding broli, you're a lovable guy, but please cut the insults, it just makes you horribly upset.
Nick
You tried aproaching some big facilities like CEA, CERN, you know the people? You want me to try and arrange an introduction ? Like I might know a friend who know's the cousin of another guy's sister in law's ex-boyfriend's nephew or something.
If you think you have an overunity device, think again, there is no such thing. You might just possibly have an unexpectedly efficient device. In which case you will be abducted by MIB and threatened by aliens.