My First Post

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ken_behrendt
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Post by ken_behrendt »

Reading through here, I had a thought, but am not sure how it could be implemented.

You guys seem to want to make a pendulum that has more weight on the downstroke than it does on the upstroke. Fine, that seems simple enough.

How about having two pendula next to each other and some arrangement whereby mass can be continuously exchanged between the two pendula? In other words, as one pendulum was beginning its upstroke, part of its mass would be transfered to the other pendulum as it began its downstroke. I'm not sure how this could be accomplished, but there would have to be some means of coupling the two pendula together so that they would remain synchronized with respect to each other at all times even though they would be out of phase with each other.

Just a thought.

ken
On 7/6/06, I found, in any overbalanced gravity wheel with rotation rate, ω, axle to CG distance d, and CG dip angle φ, the average vertical velocity of its drive weights is downward and given by:

Vaver = -2(√2)πdωcosφ
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Post by Wheeler »

Ken
This is exactly where we are going with this.
Thanks
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Post by rlortie »

Ken,

I dont see how to accomplish it either.

Two pendula one at three o'clock starting its down stroke and one at six o'clock starting it's upstoke. You want to transfer weight from six to three.

Whenever any body learns how to accomplish this let me know, I would be real interested.

Ralph
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Post by Wheeler »

Magnetism may need to be reintroduced to the pendulum at this time!
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Post by Techstuf »

Yes, and quickly.....Perhaps in speed dating format.


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re: My First Post

Post by ken_behrendt »

You guys are in luck. I just checked my virtual warehouse of past pendulum designs and found one that I previously posted and which I have again attached below.

This design uses a steel ball that rolls back and forth in a track or channel attached to the above pivot cross bar of a pendulum.

The idea is that ball rolls to the opposite side of the pivot that the pendulum weight is swinging toward. As the pendulum then begins its downstroke, its weight is effectively increased because the weight of the steel ball on the opposite rising side of the cross bar is partially reduced by the magnet above the steel ball. The process is constantly repeated everytime that the steel ball rolls to the side of the cross bar opposite to the side of the pivot on which the pendulum weight is rising.

The steel ball, however, can never reach and stick to either of the two magnets. As the section of track that the ball is on begins to rise toward the magnet, the ball is then pulled by gravity down to the other end of the track.

In a sense, this design uses some of the principles incorporated into the SMOT device to, hopefully, produce a self-pumping pendulum!

ken
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Here's a simple to construct design that might work!
Here's a simple to construct design that might work!
On 7/6/06, I found, in any overbalanced gravity wheel with rotation rate, ω, axle to CG distance d, and CG dip angle φ, the average vertical velocity of its drive weights is downward and given by:

Vaver = -2(√2)πdωcosφ
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Post by Wheeler »

Looks like the swing of thought will return to the other side again.
I like your idea of the SMOT combination and I think this may yeld some promise in the self pumping pendulum.
I hope you will let me know if it is ok with you Ken to draw or add details on your design when I have a question or idea. After all they are your designs.

I did add what I thought may cause resistance to change by magnetic attraction as the track in in it's horizonal position. What do you think?
I was thinking that the closeness of the magnetic field would tend to keep the steel ball for a period longer than desired.
I have an idea. I'll draw it and post it.
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Kens Pendulum.JPG
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Post by Fletcher »

I think you guys are 'johnny come lately's' on this one. I remember seeing the exact same thing except the steel ball went around an oval track instead of back & forwards. The magnet was supposed to lighten it but of course when it got close enough to attract it, it would hold onto it & not let go, due inverse square law etc. IIRC it is in the 'museum of unworkable devices'. If I have time I will try & find it again & post up the link.
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Post by ken_behrendt »

Wheeler...

Feel free to use my posted sketch in any way you want.

Yes, if one has the overhead magnets too close to the cross bar track, then there is the danger of the steel ball jumping off of the track and sticking to a magnet. One must consider this design to, like the actual SMOT device, require precise adjustment to get it to work. If it does work, then, with a low friction pivot, and optimized aerodynamic design, there just might be an adjustment that will always allow the pendulum to "feel" a slight pumping force acting on it during the downstroke and for part of the upstroke in either direction.

I tried to make a WM2D model of this last night, but was not successful.
This was because I needed to try to approximate the attractive forces of the magnets on the steel ball by turning on the program's "electrostatics" feature and then giving the ball and the "magnets" opposite electrostatic charges. I just could not get the effect I wanted with WM2D.

Ultimately, this design may need to actually be constructed to determine if, indeed, it is workable.


Fletcher...

I am sure that there could be a variety of variations of this design, but I think mine is probably one of the simplest and easiest to construct. If it works, then the key to making it do so will be the very careful placement of the magnets for whatever set of design factors one winds up with. I would recommend to anybody attempting construction of this device that they keep it small...perhaps no more than a foot and a half in length with a track no more than 6 inches in length.

ken
On 7/6/06, I found, in any overbalanced gravity wheel with rotation rate, ω, axle to CG distance d, and CG dip angle φ, the average vertical velocity of its drive weights is downward and given by:

Vaver = -2(√2)πdωcosφ
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Post by Jonathan »

I think this is what Fletcher means:
http://www.geocities.com/RainForest/5832/perp9.htm
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re: My First Post

Post by Mr.Umez »

About actually changing masses;

The standard theory which I have no reason to disagree with, is that the swing speed of a pendulum doesn't depend upon its mass, but its pendulum length. What lifts the pendulum back up past the bottom point is the momentum the pendulum mass has. If you eject part of the mass, you are ejecting the momentum that goes along with it. If you eject the mass, the the force due to gravity is also less. The two will nicely balance each other out so that the swing speed is the same regardless.

I still like the idea of changing the pendulum lengths during its swing. It should be easy to attach the pendulum to an odd shaped gear instead of having it swing at a fixed point. Clock-makers have their pendulums attached to a cycloid shaped gear at top to assure perfectly timed swings (hmmmm... clockmaker?)

The magnet idea is intriguing....seems like it shoud have potential? The models that Ken has shown and Jonathan linked to both have the problem the the center of mass for the ball and pendulum is always on the 'wrong' side of the swing which will cause the pendulum to quickly slow down. If, however, the magnets were placed underneath the top track, things might work out very nicely.
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Standard clock pendulum. (top part is enlarged to show idea)
Standard clock pendulum. (top part is enlarged to show idea)
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Post by rlortie »

Mr. Umez,
the swing speed is the same regardless
I am a pendulum man, therefore I am very interested in your above post. You are a true Galilean and have explained his findings that set the laws for a simple pendulum very well. This is a point that I have attempted to make for some time. It is refreshing to find there are those that understand the difference in amplitude to cycle timing and it non-relation to mass.

Your cycloid gear can be seen in an inverted position or manner on the Meresburg wheel2. jpg under drawings in the home page of this forum. The version shown is not really a gear but a slider or moving pivot. the end result lengthens the rod pivot to bob while offsetting COG just as your gear in the above post.

Your gear produces a steady, even rocking motion. The slider version throws the pivot off with gravity and centrifugal pull, the abrupt stop at end of slide transfers the now inertial force to the ascending swing. This action also shifts COG to favor the ascending side.

Regards

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re: My First Post

Post by ken_behrendt »

Jonathan...

That's the first time I have seen that device you linked to. Yes, it is definitely a variation on the design I posted. However, it seems to me that the use of a circular track with half the length of the ball's motion in either direction affected by magnetic lifting forces of the magnets is a bit too much and unnecessary. I think my approach is simpler and would yield the same results.


Mr. Umez...

You wrote:
The models that Ken has shown and Jonathan linked to both have the problem the the center of mass for the ball and pendulum is always on the 'wrong' side of the swing which will cause the pendulum to quickly slow down.
The CM of the ball/pendulum weight is only momentarily on the "wrong" side of the pivot during the downstroke, but is quickly shifted to the correct side to produce the "pumping" action on the pendulum during its upstroke. The key to making this design work is to try to reduce the weight of the steel ball as much as possible without interfering with its ability to begin rolling toward the opposite end of the cross bar channel as the pendulum swings toward the side of the cross bar on which the ball is located. One wants the ball to start rolling to the opposite end of the channel as soon as possible after the channel tilts through its horizontal orientation.

I have considered several designs that would change the length of the pendulum rod during its swings and all of these designs are complex in nature. Assuming one can make the pendulum length vary during each oscillation, then one will be able to alter the frequency of the pendulum, but this will not necessary change the height that the pendulum weight rises to on the upstroke.

ken
On 7/6/06, I found, in any overbalanced gravity wheel with rotation rate, ω, axle to CG distance d, and CG dip angle φ, the average vertical velocity of its drive weights is downward and given by:

Vaver = -2(√2)πdωcosφ
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Post by Wheeler »

I have few questions.
They may be basic, but I wondered about the following.

Would the loss of mass in the swing of the pendulum arch act indeed like that of loss of weight?

Would this loss of mass which is equal to an amount of weight act as a shorting of the length?

If the pendulum started to loose weight on the right side of the centerline, would this not slow the swing speed down?

If the pendulum started to loose weight at the center line of the arch of it's swing and continued to loose the weight at the same rate of gravitational force that is down, would this continued equal loss not make the pendulum continue to be lighter, and thus the pendulum would continue it's swing?
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Post by jim_mich »

It is not the mass of a pendulum that determines the pendulum's speed, it is the location of the pendulum's center of mass. A heavy pendulum will swing the same speed as a light pendulum IF their center of mass is the same and if their mass is shaped the same. For instance if you were to cut many pendulums (and rods) out of sheet metal. Each would swing the same. Stack them all together and they will still swing the same.

Now if you were to have a single rod or cord holding a long skinny horizontal rod shaped pendulum and compared its swing with the pendulum oriented in different directions then the swings will be different. When the rod is oriented so you only see the small pendulum end it will swing faster. If the pendulum is turned 90 degrees so you are looking at the side of the pendulum then the outer ends will need to swing faster because they are farther from the pivot, which will make the whole pendulum swing slower. This is the same reason that a solid wood cylinder will roll downhill faster than a metal tube that is the same weight. Both recieve the same gavity pull because they weigh the same, but the metal tube has a higher percent of weight out at the edge which must be accelerated.

Droping a chunk out of the weight would change the swing ONLY if it changed the weight's center of gravity.

The bottom line is that a pendulum's shape and distribution of weight effect it's swing, but not it's total weight.

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