Well, if you don’t mind, I’ll stick with my diagram, since I had a hard time making anything out on yours.

Some hints you may appreciate:

Those little blobs on the end of the leaf springs aren’t accidents. At one end they act as catches, preventing the BB from rolling back down the slope, while at the other they become a trigger, when the tension on the leaf spring bends it far enough, the blob will be rotated out of the way of the BB, causing the mechanism to fire (for speed, I just used a dot on the end, in reality it would be better to use a standard pawl shape). This inbuilt trigger should cause the BB to fire at a consistent point on the cycle, whereas relying on friction would not; an early misfire would obviously leave the ball at the wrong end of the travelling arm and the wrong side of the leaf spring, breaking the cycle.

The leaf spring is deliberately shown bending through the travelling arm. If the LS were completely contained in the arm, its distortion may actually trap the ball, causing it to release too late, if at all. A better design is to leave the LS unconstrained along at least part of its length, which can be accomplished via a small open slot on one side of the travelling arm.

And some you will not:

The active weights will of necessity have to be much heavier that the BBs, otherwise they will not tension the leaf spring sufficiently to fire the BB. Similarly the leaf spring must be stiff enough or it will reach its maximum deformation without storing enough elastic potential energy to launch the BB to the other end. These two facts have an important consequence.

If you built your design with weights and springs capable of firing the BB from the 8 to the (just past) 2 o’clock positions then set it up with one arm exactly vertically, you would see that the wheel appears unbalanced by the presence a BB at the 3 o’clock position that is not present at the 9. However, looking closer, you would also see that the much heavier active weights at the 7 and 8 o’clock positions are raised slightly higher that those at the 4 and 5. This would create a sight imbalance in the opposite direction.

Using my version of your mechanism as a guide, I calculated that an active weight of about 40 times the mass of the BB (the actual number depends on the stiffness of the leaf spring, which I didn’t bother to work out) would be sufficient to launch it the full distance, but that then the opposing moment of the elevated weights was greater that the intended moment generated by the BB. In short, the wheel would slow with every BB fired and eventually stop.

It is allot of fun to debate with people who specialize in reporting and debunking hoaxes

I don’t specialise in it. But I did do an engineering degree that included a healthy dose of statics and dynamics. Analysis by which does not support your claim.

When you build a working one, I’ll eat my words. But until then, I’m afraid, you’re a loony!

P.S. Whether it works or not, if you built a brass ‘steampunk’ version, I’d buy one!

David B.

Well that makes Loony my hobby. LOL

Yes I saw the knob on the end of the spring, and I thought to my self that he was truly thinking, but you are wrong on the 9:00 position. The shift has to happen before the equilibrium mark to keep it from slowing down, which is 8:30. Like I said in earlier post that this is just one of the shooters, but I will tell you about the effects of another, which flings the disk, but the reason I didn’t draw it in, 1 it is one if my hush hush ones, and for I was trying to figure out Besslers 12 ft wheel which went both directions and the leaf spring seemed to be the only one so far for a bye directional. The disappointments I have learned with this one is it wouldn’t go that fast, so posting it would not be any danger to what I am working with now. But I am building a 4 chamber version for the other shooter which should make for a good toy.

Well I got to go.

I was thinking about where the ball would be in each of the six arms of your design as it was in motion.  Here’s a little diagram; each image shows the estimated position of the ball in one of the six arms of the wheel at one particular instant in time.

In arm number (1), the ball is still in the leaf-spring, just before being launched.  In (2), the ball is bouncing around at random in the arm; it could be on the left side, it could be on the right side, it could be in the middle, it could be in mid-air, it could have hit the axle and bounced straight back to where it was launched from. . .there’s no way to tell where it will be pushing down on the arm, or if it even will be (if it’s in mid-air, it wouldn’t be having any effect at all).  In arm (3), the ball will either be in the lower end or on its way there.  In arms (4), (5), and (6) it will be down in the bottom of the arms.

For every arm in which the ball is to the right of the axle, the wheel gains force in its spin.  For every arm in which the ball is to the left of the axle, its spin loses force.  If the ball is in mid-air, then that arm has no effect at all on the force.

The amount of force the ball adds to the whole system depends on how high the ball is as well as how far out from the axle it is.  A ball at the five o’clock position would add a lot less force to the whole system than would a ball at the three o’clock position.  Furthermore, a ball that is all the way at the end of the arm at the three o’clock position would give more force than would a ball that was only halfway to the end of the arm at the three o’clock position.  There are lots of fun physics equations showing all this (PE = mgh for the former case, another one about torque that I don’t have all the symbols to type out for the latter case).

So keeping that in mind, look back at the first diagram of where the ball is in each of the arms at a given moment.  Not counting arm (2), in which we don’t know where the ball is, we’ll have an overall net force against the spinning of the wheel.  Arms (4) and (6) cancel each other out, while arm (5) isn’t doing much of anything at all.  Arm (3) partially cancels out arm (1), but probably won’t do so entirely. . .which leaves a small bit of the force in arm (1) acting against the spin of your wheel.  So the ball in arm (2) is the critical one, and that ball can be anywhere.  Arm (3) is also a bit of a tricky one, but less so than is arm (2).

Momentum will be both helping and hindering you.  It will help you in that once you get the wheel spinning, it will have a tendency to keep on spinning a bit even with no additional force acting on it.  This can help you overcome some occasional minor forces acting against it, and might give the ball in arm (6) some slight ability in giving you some positive force in that position rather than it being a totally neutral position.  On the other hand, angular momentum will mean that in arm (2) the ball will be trying to force its way out through the end of the arm and away from the center of the wheel, which is pretty much opposite of the direction that you want it to go when you launch it.  So you’ll have that angular momentum of the ball to overcome when you launch it.  The faster the wheel spins, the more angular momentum the spring will have to overcome.  How much this will count against you will obviously depend on a lot of factors, such as the mass of the ball, the radii of the arms, and the force given to the ball at launch.  But there will be a rate of spin at and above which the wheel will stop working.

Also, the action of your end weights will be acting against you.  Between the twelve o’clock and the one or two o’clock positions, one of the weights will very gently come to rest against the arm.  This will be a fairly gradual event, so it won’t provide much if any force (if it did, it would be force that would be in favour of your wheel spinning).  At the eight o’clock position, though, you have your spring launch the ball and the end weight suddenly slamming down onto the arm.  This is not a gradual movement like at the one o’clock position.  It will act against the spin of your wheel.  So every time your wheel launches one of its balls, it will loose energy that it doesn’t gain back in the opposite position.

Plus, you still have friction with the axle to deal with.

Tah - 11 January 2008 11:40 PM

As with everyone else here, I think (know) that your machine won’t work due to the various problems and issues already pointed out.

But if you enjoy doing it and want to work on it you go right ahead.  More power to you.  It would be awesome for you to accomplish a perpetual motion machine.  But none of us are holding our breath.

And when you run into problems with it feel free to come here and ask for comments and advice.  As you can see, there are enough people here who know what they’re talking about that they’ll be able to point out where the errors are in your calculations and assumptions.

Oh, sure.  I’d be absolutely delighted if somebody managed to make a perpetual motion machine that actually was one.  It would do all sorts of interesting things to the world of physics.  And if somebody managed to work out how to make one that actually provided energy. . .well, that would have no end of beneficial uses.  It’s not like I’d be angry if he was successful.

The problem is that there are a lot of people who pretend to make one and lie about it, and many others who genuinely believe that they have made one but are wrong.  And many of the latter simply refuse to accept the truth, even when it’s plainly shown to them.  I can’t actually conclusively show the device to be workable or not, though, unless I see a real working model and can get real measurements for all the forces involved.

Mainly, I’m just pointing out what potential problems I can see in AB Hammer’s ideas, and seeing if and how he’s already addressed them.  For one thing, if we find some serious error then it will save him the effort of building the whole device wrong.  Besides, it’s just fun to poke and pick at the idea to see what I can find and figure out. 

First a mea culpa. I said:

David B. - 12 January 2008 12:24 PM

Using my version of your mechanism as a guide, I calculated that an active weight of about 40 times the mass of the BB (the actual number depends on the stiffness of the leaf spring, which I didn’t bother to work out).

1) That should be 20 times the mass of the BB, not 40 times, I neglected to take into account that as the weight is effectively pivoting at one end, the centre of mass is only raised by about half as much as the swinging end is[*].

2) It doesn’t depend on the stiffness of the leaf spring, as long as the spring is sufficiently stiff not to fully deform (i.e. bend until it meets the bottom of the arm), it doesn’t really matter how stiff it is.

That said…

AB Hammer - 12 January 2008 03:29 PM

but you are wrong on the 9:00 position. The shift has to happen before the equilibrium mark to keep it from slowing down, which is 8:30.

My only comment about the 9:00 position was:

[T]he wheel appears unbalanced by the presence a BB at the 3 o’clock position that is not present at the 9.

This is clearly true in both my and your drawings. I assumed the device in my diagram would be tuned to fire the BB at 8:00:01, i.e. I analysed in on the very cusp of firing the BB. This analysis clearly showed that the equilibrium position at that time was anticlockwise of the 8:00 position, hence the device would be slowing at that point. That there is an equilibrium position for the design (with the ‘firing’ arm at approx 07:50, though I haven’t bothered to work it out exactly), shows that the device is quite capable of reaching an equilibrium.

Just a thought, AB, but why don’t you make the whole device out of brass or ally except for the BBs? That way you could fix a strong magnet underneath the 4:00 to 5:00 position to attract the steel BB downwards and get even more power out of your device?

[* This does not make a difference to the final outcome, though the device would slow a lot less rapidly than I originally thought.]

As ever in life, the rule here is ‘follow the money’, or in this case ‘energy’.

As an arm on AB’s device moves from the 6:00 position to the 8:00 firing position, the weight will try to fall under gravity. It’s motion will be resisted by the spring and the distance its centre of mass does fall x mass x g (i.e. UG) will be slightly more than the elastic potential energy (UE) stored in the spring (the difference will go into heating the spring slightly, which happens to metal when you bend it).

When the device fires, the kinetic energy given to the ball (EK) is slightly less than the UE of the spring, with the difference this time dissipating as sound (the ‘click’) and a bit more heat.

Ignoring the losses to friction of the BB’s path up the arm (the friction of ball bearings on metal is quite low, that’s why we use them as bearings), the EK is turned into gravitational potential energy again (UG2) as the ball moves up the slope, slowing as it goes.

In fact not only is EK greater that UG2, it is significantly greater. The ball must be still moving fast enough at the far end of its path to be able to move the catch mechanism (spring pawl or whatever) or it’ll bounce off and return to the firing end, but the wrong side of the leaf spring. This means that not only is more energy lost in heating the spring, but that the ball will travel slightly past the catch point, fall back, then come to rest, wasting that excess EK. Also, if the ball is moving too fast, more excess EK will be lost when the ball hits the far end, and if excessively fast and it will rebound with enough force to overcome the catch and return to the spent firing end.

Hence we have UG > UE > EK > UG2, an energy loss over the course of the BB’s path from the 8:00 to 2:xx position (n.b. the slower the ball is moving, the further the far end of the arm will have rotated from 2:00 by the time the BB gets there).

As the weight must be raised back to it’s intial position at a cost of UG, but the ball will only release UG2 in returning to it’s previous position, this represents a loss for the entire machine.

[Edit: Cleaned up explanation and made it a bit more readable]

Accipiter - 12 January 2008 11:25 PM

Also, the action of your end weights will be acting against you. [...] At the eight o’clock position, though, you have your spring launch the ball and the end weight suddenly slamming down onto the arm.  This is not a gradual movement like at the one o’clock position.  It will act against the spin of your wheel.

Acci, I’m surprised at you. Imagine sitting on one end of a seesaw with an equal weight on the opposite end. No matter how hard or repeatedly [**] you hammer at your end you will not be able to drive it down to the ground (unless you bend/break it). In fact, if you think carefully about it, every time you bring the hammer down, your end of the seesaw will come up slightly to meet it[*]. The action of the falling weights would be in favour of the spin of the wheel, not against!

[* As you pull down on the hammer, the hammer also pulls up on you, hence the force of you (and the hammer) on the seesaw is reduced (by the approximately the force you are applying to the hammer), hence no longer compensates for the force from the counterweight, causing your end to rise.]

[** Correction: if you repeatedly hammer at one end, you can of course cause your end to drop to the ground. You simply raise your hammer in time with each dip of the arm and drop it in time with each rise, by continually doing so in resonance with the motion of the seesaw you should be able to increase the amplitude of the rocking motion until your end touches down. This principle is known to most schoolchildren who have ever ridden on a swing, but only occurred to me 3 seconds after I had logged out. Nuts!]

Greeting Guys

I thought I would start by giving you a link to information on Bessler. In that sight you will find 143 drawings known as MTs. These come from a book that Bessler wrote getting ready to teach about Perpetual Motion. None of the MTs work even though you can look at them and wonder why knot.

Here is the link
http://www.orffyre.com/main.html

  I figure this will help out some of your thoughts, for the icon one is a derivative of MT137 which only shows the chambers.
I will go ahead and build an 4 chamber version with the single directional shooter it only requires 5 times the weight of the disk or ball and it has a really cool catch, as soon as the disk or ball hits the other side a pen comes down and catches the disk or ball, but when it gets to the shooting location it drops when the shooter shoots, (they are connected together). I am starting to feel that the pinch spring even when worked out, won’t be dependable for I didn’t put in the heat factor which will cause the spring to loose its capabilities. But as an inventor I will always have at least 3 back-up design. That is an unwritten rule that I go by.

When you file your patent, you might want to think of using a spell checker.

David B.

Just a thought, AB, but why don

OOPs

Corrected