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A disc rotating at a certain rpm (e.g. at 1000rpm) having a radial hole with a plunger or piston inside the hole. Say the mass of this piston is 1kg. Certainly this piston will move radially outward under the action of centrifugal force. What will be the speed or velocity of this piston in the radial direction? Or how much distance it can travel radially outward in 1 revolution of the disc?
The piston would be subjected to an acceleration of, where ω is the rotational velocity of the wheel in radians/second and r is the piston's position measured from the center of the wheel. This assumes that ω is constant and that there is no friction resisting the motion of the piston. From this equation you can see that as the piston slides outward it is subjected to an ever greater radial acceleration, so its speed will increase dramatically. The equation of motion for the piston will be of the form:
Whether the piston slides out of the wheel one rotation or not depends on the initial position of the pistonand the radius of the wheel. Note that the mass of the piston is immaterial here.
**EDIT** See improved solution in the following post
Update: upon refelection I realized the solution is a bit different from what I originally wrote. The differential equation for the motion of the piston is:
Asumingis constant, the general solution for this is:
where A and B are constants, whose values can be determined from boundary conditions - namely the initial position and initial radial velocity of the piston at time t = 0. For example, if the piston is originally at position, and its initial velocity is 0, then you get:
From this the solution is:
Thank you for your consideration. As an example, suppose that wheel is rotating at 1500 rpm, piston is initially at 100 mm from the wheel axis of rotation.
Q.1 How much distance the piston will move outward in 1 revolution of the wheel while rotating at 1500 rpm?
Q.2 If the piston has mass of 1 kg, it will exert centrifugal force of about 251 kg force at 100 mm radius. Is it correct? Please verify.
Q1.: Using the formula I gave you, you need to convert 1500 RPM into the equivalent values for ω and the time taken for one revolution, T:
Soor one revolution. Hence assuming there is no friction, the distance the piston slides in one revolution is:
Q2: when the piston is at r=0.1m, its centripedal acceleration is:
And the force exerted is:
If the movement of the piston is opposed say by some sort of hydraulic or spring force equal to half of the centrifugal force (i.e. 2467N), then what will be the speed of the piston at that particular point?
If the spring is designed so that its opposing force is half of the force due to centripedal acceleration regardless of the pistons radial position, then the pistons radial acceleration is half of what I provided earlier. Hence its velocity at any point in time is half of what it would be without the spring.Quote:
Originally Posted by mayya
But if the spring is designed to provide 2467N no matter what position the piston is in, then theoretically the piston is in equilibrium in its starting position of 0.100 m, so it doesn't move at all. However this is an unstable equilibrium, so the piston will actually either move inward or outward - it's impossible to tell. It's like balancing a pencil on its point and trying to figure out which way will it fall.
Mayya - perhaps you can explain why you're asking these questions - what exactly are you trying to design?
Thanks for your replies as they helped me a lot. Actually I am asking these questions on behalf of my father who is trying to design a mechanism for utilization of centrifugal force. With thanks and anticipation.
It can be confirmed by calculation that the piston having 1kg mass with its centre of gravity at 100 mm while rotating at 1500 rpm needs about 185 watts of input power/energy. Whereas it is generating a centrifugal force of 2467N just calculated by you. If this radial force can be diverted somehow and applied tangentially to drive another disk having 100 mm radius, it can produce tremendous power (about 19 kilowatt).
With reference to the same data, what will be the velocity of the piston in radial direction at particular distance from the center e.g. at 200 mm radius?
mayya - are you asking about the velocity of the unrestrained piston (no opposing force)? Go back to post #3, which gives the radial position of the piston as a function of time:
If you set R_0 = 100 mm and use a rotational velocity of 1500 RPM = 157 rad/s, then the elapsed time for the piston to reach position r = 200 mm:
The radial velocity is the derivative of the position:
Substitute in the value for t that we found above:
Note that all this assumes that the wheel is rotating at a constant speed of 1500 RPM. Hope this helps.
I'd like to see your calculation of this. It seems like you are thinking that you could build a perpetual motion machine, that with input power of 185 watts it could generate 19 kilowatts of output. Of course that's impossible, at least for continuous operation.Quote:
Originally Posted by mayya
Many thanks for your helping efforts.
Regarding your question, reason for consideration of centrifugal force for free power generation is due to following calculations and logics.
Rotation of one kilogram mass with its center of gravity at 100 mm from axis of rotation at 1500 rpm (25rps) requires input power equal to its total kinetic energy. Using Steiner's parallel axis theorem this comes out to about 185 joules/sec (watt) and the tangential force involved for torque is just 11.78 N. On the other hand centrifugal force naturally created in this process is 2467 N in radial direction along with acceleration of 2467 m/s2. If this much force may be diverted to be applied tangentially at 50 mm radius to produce torque at 1500 rpm the power output comes out to be about 19379 joules/sec (watt), 104 times the input! With increase in radius output will increase proportionally. Keeping in view practical resistance/constraints whatsoever, if only 50% (52 times) or even 10% (10 times) of that is achieved then it may be a wonderful revolution.
A dynamic (moving) force always does the useful work. If the force is dead, that is not moving like hydraulic pressure force in a pressurized container, it does produce stresses but no useful work is done unless the pressurized fluid is moving. In spite of great confusion about reality of centrifugal force, it is a real, dynamic and feel-able force equal and opposite to centripetal force having same acceleration and velocity in magnitude but opposite in direction. That is why the piston under discussion travels radially outward with velocity yet to be analyzed. Suction of oil from the reservoir quite below the hydraulic pump, particularly of radial piston type, due to centrifugal force of outwardly extending pistons is another indication of reality and usefulness of centrifugal force.
Other than the torque that may be produced or not, radial travel of piston under discussion through certain distance with that much dynamic force at the expense of very little input force/energy, can do too much work accordingly. Certainly, continuous working and harness of centrifugal force with this sort of arrangement is impossible. But there may be other possible options to be discussed later. Oscillations produced by centrifugal force with surplus energy have already been worked upon a lot in manual way by Veljko Milković. The continuous motorized oscillation converted into rotation has been experienced and being tested myself in a different way. Seemingly more viable one wherein much of the peaks of radial centrifugal force may be directly converted into continuous torque/circular-motion requires authentication of radial velocity of radially moving mass under the influence of centrifugal force at certain radius. Working of perpetual motion machine or free energy generation just depends upon the fact whether radial velocity is how much more or less than tangential velocity at particular radius.
Now coming back to my questions and your calculation in this regard the following may kindly be reviewed.
Firstly, calculated distance traveled by the piston is about 27 (=26.77) meters radially outward in one revolution. This is covered in 1/25 seconds as there are 25 revolutions per second (1500 rpm). Therefore, average velocity/speed would be (27 x 25 =) 675 m/s.
Secondly, acceleration (a) at 100 mm radius (r) calculated above is 2467 m/s2 then acceleration at 200 mm radius would be 4934 as value of acceleration has been doubled.. If we start at the center, initial velocity/acceleration is zero then, radial velocity at 100mm & 200mm would be 2476 & 4934 m/s respectively and after one second time it would become quite higher further in proportion to radius occupied by that time.
Thirdly, calculated time taken by piston from 100 mm to reach 200 mm is 0.0083 second. Then average velocity at 200 mm radius would be acceleration x time = 4934 x 0.0083 = 41 meter.
Fourthly, quite big value of acceleration as well as the heavy centrifugal force as compared to tangential force involved for rotation of I kg mass at any radius, apart from complex calculus, logically indicates larger values of radial velocity increasing with increasing radius.
All above needs verification/better understanding of which value of average radial velocity/speed of piston, 27, 675 or 4934 m/s is correct.
It is quite clear that tangential velocity at any radius is equal to rotational velocity at that point converted into linear dimensions. If piston escapes rotation at any radius it will leave tangentially with the same velocity as centrifugal force disappears there instantly.
Kindly don't mind my poor understanding of complex mathematics/calculus as I need very simple and clear comparison of rotational/tangential and radial velocities at any radius. In spite of serious search I could not find relevant information either in available books or on the net.
Thanks and regards.
You are confusing torque, energy, and power. This device has angular kinetic energy of 123 N-m . To maintain the rotational velocity of 1500 RPM actually requires no power at all - as long as there is not any friction nor any resisting force.Quote:
Originally Posted by mayya
Again - force is NOT power. Please show me your calculation of this so I can comment on the error you are making here.Quote:
Originally Posted by mayya
This is the first time you have mentioned a resistant force - namely sucking oil from some sort of reservoir. The problem is that the resistant force of the oil causes your piston to move MUCH slower radially than previously calculated. The work done by the priston is the force being applied by it (the 2467N figure) times the radial velocity of the piston, which is yet to be determined.Quote:
Originally Posted by mayya
Wrong. As I've said before this is true ONLY if there is no resistive force (i.e the piston does no work). Once you introduce the resistive force of the oil, so that the piston is actually doing some work, this velocity is greatly reduced.Quote:
Originally Posted by mayya
Yes, but only if there is no resistive force (how many times do I have to say this? )Quote:
Originally Posted by mayya
None of these values are correct for the machine you've proposed.Quote:
Originally Posted by mayya
The key here is that if the piston does work against the oil reservoir (and ultimately in turning a generator) its kinetic energy is decreased by an equivalent amount. This means its radial velocity decreases as well. In the extreme, if the radial velocity is 0 then the piston can do no work at all, EVEN THOUGH IT'S TANGENTIAL VELOCITY IS STILL QUITE HIGH. If you try to convert that tangental velocity into work, the machine will slow down, or else require additional power input to keep turning.Quote:
Originally Posted by mayya
There is nothing here that suggests any sort of "break through" with respect to designing a perpetual motion machine. Thinking through the physics of a machine such as this can is a good intellectual exercise that can help you build a better undertanding of how work and energy are traded off against each other. But one thing is for sure: if your conclusion is that you can design a machine that produces more energy than it consumes, you have made an error in your thinking.
A lot of mix-up and confusion has taken place. I shall show you detailed calculation a little bit later. Actually, I expected calculation based on this data yourself and confirmation of my calculations thereby.
I have, though may not be perfect, quite sufficient concept of force, torque, work, energy and power etc. All discussion pertains to same data from the very beginning under ideal conditions. I did never mention motion of piston in radial direction under the influence of centrifugal force against any resistance. Example of intake of oil by the piston was quoted just to emphasize the existence, dynamicity and ability of centrifugal force to work. As you mentioned earlier, distance and velocity of radial travel of piston against any resistance will certainly decrease in proportion to the resistance and become zero for 2467 N. Also, I am not talking about extracting centrifugal force energy of the piston mass for free power generation. It is quite a different story and mechanism to be discussed in detail later. Of course, over unity machine is impossible but excitation and incorporation of any natural energy cannot be denied.
Question at the moment is just to explore exact and confirm value of radial velocity of piston or any radially slid-able mass under the influence of centrifugal force at certain radius. This will help analytical comparison of the same with tangential velocity (= rotational velocity in linear dimensions) at certain radius whereas practical confirmation needs a lot of R & D and investment. What I have observed physically through my practical experiments, radial velocity obviously seems to be much higher than tangential velocity at certain radius. Relevant values calculated or estimated logically so far are confusing as they do not tally each other as mentioned in my previous query and reproduced hereunder.
“Firstly, calculated distance traveled by the piston is about 27 (=26.77) meters radially outward in one revolution. This is covered in 1/25 seconds as there are 25 revolutions per second (1500 rpm). Therefore, average velocity/speed would be (27 x 25 =) 675 m/s.
Secondly, acceleration (a) at 100 mm radius (r) calculated above is 2467 m/s2 then acceleration at 200 mm radius would be 4934 as value of radius has been doubled.. If we start at the center, initial velocity/acceleration is zero then, radial velocity at 100mm & 200mm would be 2476 & 4934 m/s respectively and after one second time it would become quite higher further in proportion to radius occupied by that time.
Thirdly, calculated time taken by piston from 100 mm to reach 200 mm is 0.0083 second. Then average velocity at 200 mm radius would be acceleration x time = 4934 x 0.0083 = 41 meter.
Fourthly, quite big value of acceleration as well as the heavy centrifugal force as compared to tangential force involved for rotation of I kg mass at any radius, apart from complex calculus, logically indicates larger values of radial velocity increasing with increasing radius.
All above needs verification/better understanding of which value of average radial velocity/speed of piston, 27, 675 or 4934 m/s or even else is correct.”
Yes, that would be the average veocity. But the actual velocity at t = 0.04 seconds turns out to be 4205 m/s.Quote:
Originally Posted by mayya
Yes - clearly as time elapses and the piston slides further outward its acceleration increases exponentially. In fact, given an ideal piston with no friction forces, IF the disc is large enouh and IF you could actually maintain 1500 RPM as the piston slides outward then believe it or not within 1 second the piston would be moving FASTER THAN THE SPEED OF LIGHT! Clearly there is an issue here - see below.Quote:
Originally Posted by mayya
Acceleration is not a constant, so your average calculation doesn't mean very much. As already noted, the actual velocity for r = 200mm is 2467 m/s, and the actual accelaration for r = 200 mm is 4934 m/s^2Quote:
Originally Posted by mayya
Yes, but... As noted below, you have set up an impossible task.Quote:
Originally Posted by mayya
The term "average velocity" really has no meaning.Quote:
Originally Posted by mayya
Agree. It's a great way to make a catapult.Quote:
Originally Posted by mayya
As I have noted in previous posts, the formulas that govern the motion of the piston under ideal conditions where (a) the rotational veocity is constant 1500RPM, (b) the piston starts at r = 100mm, and (c) there is no friction, is as follows:Quote:
Originally Posted by mayya
Note that these are all exponential functions - and as I will show below the numbers become outrageously large very quickly. But what really makes this whole project impossible is that it assumes that the disc can be spun at a constant rate of 1500 RPM, even as the piston slides outward. But this isi bver y difficult, and quickly becomes absolutely impossible as the piston slides further out. The moment of inertia of the piston increases as it slides outward, and so to maintain a constant spin rate more power is needed to turn the disc. This is because in addition to the radial velocity and acceleration that we calculated earlier, there is also a tangential acceleration component, which in turn introduces correolis forces into the mechanism. The way it works is this - the tangential acceleration of the piston is:
The force that the piston applies to the side walls of its cylinder is therefore:
and the torque and power required to keep the disc spinning at a constant RPM is:
Now, to crunch some numbers: at t = 0.0083 seconds you get:
r =0.2m
v = 27.2 m/s
a_r = 4934 m/s^2
a_t = 8547 m/s^2
T = 1709 N-m
P = 268522 watts
You see the problem? You need an impossibly large motor to keep this mechanism spinning at 1500 RPM. To take it ridiculous extremes - using these classical physics calculations, if you wanted a disc that was of sufficient size for the piston to take 1 second to reach the edge, the disc would need to be 8.3 x 10^66 meters across! And the piston's velocity at one second would be 1.3 x 10^69 m/s! Obviously this calculation is in error as it does not take into account the effects of relativity. In brief as the piston approaches the speed of light its mass would increases to near infinity and the power required to keep the disc turning becomes infinite as well.
Bottom line - it is impractical for the mechanism to maintain a constant RPM as the piston slides outward.
Mayya - these graphs show the piston's position, velocity and acceleration for the first 0.0083 seconds. Note that the tangential acceleration becomes huge - and hence the power needed to keep the disc turning becomes huge as well.
My assumption and calculation regarding amplification of input energy by utilizing centrifugal force are as under for your review and comments.
Consider a body having one kilogram mass (m) with its center of gravity at 0.10 meter radius (r) from axis rotating at 1500 rpm (25rps). It will require work input (Wip) or energy equal to its total kinetic energy comprising kinetic energy of translation plus kinetic energy of rotation.
Wip = ½ x mass x square of linear velocity of center of gravity + ½ mass moment of
inertia (I) x rotational velocity (ω)
= ½ mv^2 + ½ I ω^2
= ½ m(2 π.r.n)^2 + ½(½ m.r^2)(2 π.n)^2
= 2m(π .r.n)^2 + m(π.r.n)^2
= 3m(π.r.n)^2 Nm, joules
When n is in number of revolutions per second then power input (Pip) becomes
Pip = 3m(π.r.n)^2 Nm/s, joules/s, watt
= 3 x 1.0(π x 0.10 x 25)^2
= 185 watt
As power equals to torque (T) x rotational velocity and torque equals force (F) x radius,
Pip = Tip.ω and Tip = Fip.r
So Fip = Pip/r.ω
= 3m(π.r.n)^2/r.ω
= 3m(π.r.n)^2/2 π.r.n
= 3/2m. π .r.n
= 3/2 x 1.0 x π x 0.10 x 25
= 11.78 N
The centrifugal force naturally created in this process, I would like to call it the output force (Fop), at the expense of 11.78 N will be
Fop = m.r.ω^2
= m.r(2π.n)^2
= 4 m.r(π.n)^2
= 4 x 1.0 x 0.10(π x 25)^2
= 2467.40 N
This output centrifugal force (Fop) has also acceleration of 2467.40 m/s2 directed radially outward.
If this much force having big acceleration as well may somehow be applied tangentially at 50 mm radius to produce torque at 1500 rpm the power output (Pop) comes out to be
Pop = T.ω
= Fop.r.ω
= 4 m.r(π.n)^2. r. 2 π.n
= 8 m.r^2. π^3.n^3
= 8 x 1.0 x 0.05^2 x π^3 x 25^3
= 19378.92 watt
If Fop is applied at 100 mm radius
Pop = 8 x 1.0 x 0.10^2 x π^3 x 25^3
= 38757.84 watt.
Thus the ratio output force to input force becomes
Fop/Fip = 2467.40/11.78
= 104.75
And ratio of output power to input power becomes
Pop/ Pip = 19378.92/185
= 104.75 for 50 mm radius.
and = 38757.84/185
= 209.5 for 100 mm radius
and similarly varies in direct proportion to value of radius.
It may be reminded here that I am not talking about extraction of centrifugal force energy of radially sliding piston mass for free power generation. It is quite a different mechanism to be discussed in detail later. The case of piston has been quoted just to analyze the radial velocity of piston or any radially slid-able mass under the influence of centrifugal force at certain radius.
As you agreed to most of my point raised in last posting, high values of centrifugal force associated with huge acceleration at even smaller radii is also suggestive of high radial velocity accordingly as compared to tangential linear velocity at relevant radius. So piston can still move against reasonable resistance to give output work/energy at velocity comparable with tangential velocity.
Keeping all the complex calculus aside, let us consider just one simple logic for the time being. Suppose the piston initially occupies a position with its center of gravity at zero radius or at any minimum (a few millimeter) radius but prevented to slide outward while the disc is rotating at its speed of 1500 rpm. So, initial acceleration and velocity of piston will be zero. Then piton is suddenly released to slide. In just fraction of a second it will reach 50 or 100 mm radius with centrifugal force and acceleration as calculated above. Then velocity just after a fraction of second or after one complete second will be equal to the value of acceleration by that time which comes out to be much higher than relevant tangential velocities there. Is it correct or not?
It is noted that query prepared in word document when posted does not show some of the characters or operators like omega, pi, etc. in actual format in relevant formulae. Kindly advise what to do to solve this problem.
It is noted that query prepared in word document when posted does not show some of the characters or operators like omega, pi, etc. in actual format in relevant formulae. Kindly advise what to do to solve this problem.
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