BesslerWheel.com

[pequaide]

If you apply a force on the end of a lever the force is multiplied as you approach the fulcrum. If you apply 5 newtons at one meter you are apply 10 newtons at ½ meter. At 1/10 meter you are applying 50 newtons.

Now let’s place 5 kilograms of inertia on the end of the one meter lever. If we apply force at ½ meter the five kilograms will act as if it is 10 kg. And if we apply force at 1/10 meter the 5 kilograms will act as if it is 50 kilograms. Or 1.43 kilograms at 11 inch acts like 21 kilograms at .75 inches.

[pequaide]

I did not have some parts that I needed and I had some computer time. So I looked up 'law of the lever' in wikipedia.

My understanding of the law is exactly like their definition and all my last experiment (21 at .75 and 1.43 at 11) does is apply and prove the law of the lever. Especially interesting is their use of the word power.

The input forces (engine) and drag forces (flaps) on an aircraft wing would all be law of lever forces, so this has to be one of the most applied and proven laws in the world of engineering.

Why then: when you are properly applying a law, do you get attached by people on this site? Words used like delusional, fraudulent, needs his frontal lobe stimulated , has him nailed, I will drink to that, etc etc.

I am making a short list of non-read posters. So if you see a non-response to a question (probably an insult), I will give no response because I hopefully did not read it. This will save time and reduce my level of irritation. I have simple wasted to much time on people that 'don't think I can do an experiment'.

Where does that leave the project? Well the next step is to leave the massive flywheel out of the Atwood's. A request that could have been made without the insults.

I am dropping down to a ¾ inch shaft spinning a 19 inch plywood wheel. I did not choose to alter the throwing 48 kilogram flywheel. And the bigger bearings are stiff of course. I think the photo gates could pick up a temperature change in the room. The lubricant in the bearing would get a bit stiffer for different reading on different days. The reading are consistently in the same area but different at different temperatures. At least that is my guess.

[pequaide]

I just realized something.

I increased the circumference mass for another throw on 48blue.

So I increased the circumference mass from 1.43 kg * 2 = 2.86 kg; to 2.28 kg * 2 = 4.56 kg. And I left 42 kilograms (42 kg balanced and 7.5 lbs unbalanced) at .75 inches. I did not expect much of a slow down in acceleration because the mass increase (mr) is only about 3%. And there was not much slow down; the time was a little over 7seconds at 7.12 sec, for a final velocity of about .277 m/sec. Before it was .28 m/sec so it was pretty much what I expected.

What I realized was; what if the formula was mrr?

It would not have worked; the slow down would have been much greater.

It is also clear that it is hard to slow this thing down. You can get a good deal of motion for a small mass drop.

[pequaide]

I was throwing with 48blue. The slow throws have to release the missile from the surface around 9 o'clock when throwing counter clockwise. I can throw the tethered 2.28 kilograms over the top but the release point sends the tether and missile into the ground. The photo gates can not get anywhere near this because a mistake clears the scene of any obstructions. A miss throw is now breaking 1/16 inch nylon cord. I could harden the sight for the photo gates or just forget about the slow throw and play hard ball. But I sat on the sofa and reviewed the process in its simplest form. The double pulley is another example of how Atwood's work at different radii. And the cylinder and spheres with the 90 degree stop shows that a wheel throws momentum not energy. So here we go again in simplest form.

An Atwood’s can accelerate 10 kilograms to 1 meter per second or 1 kilogram at 10 meters per second with the same amount of drop. It is the same amount of drop for the same amount of momentum: not energy.

A wheel will throw a tethered mass from it circumference; and when it does, it throws linear Newtonian momentum: not energy.

From the double radii wheel I throw a 98 gram missile (BB bag) on a 52 inch tether from the 18 inch diameter and it stops the wheel nicely. My taped on pin, which was an Allen wrench, fell of so I switched to the 12 inch diameter to see what would happen. The 98 gram missile on the 52 inch tether stopped the wheel just as nicely from the 12 inch diameter as it did from the 18 inch diameter. The tether unwrapped about another half rotation and threw in a different direction but the wheel stopped beautifully.

Not only does the wheel throw momentum it does not care what radius it throws from.

Apparently the one third less leverage of the 12 inch diameter (compared to the 18 inch diameter) is compensated for by the one third longer unwind. The tether is about .92 times the length of the 18 inch circumference and about 1.38 times the length of the 12 inch circumference.

[pequaide]

Here is some more interesting data.

I increased the tether length to a full circumference for the 18 inch diameter wheel, about 56 inches. When this 56 inch tether is (weighted with 98.2 grams) flung out from the wheel it will stop the wheel with a very slight backward motion. This backward motion means the 98.2 g is just a tiny bit to heavy. But BB bag are hard to make so I will accept the slight backward motion.

The 56inch 98 gram tether will stop the wheel in the same manner from the 12 diameter or the 18 inch diameter. The double wheel (or sometimes used as a double pulley) is pictured on page 80.

Then I got to thinking: With one full tether wrap of the 12 inch wheel the mass of the BB bag would have to be around 150 grams. So I did a blind study. I when to my war chest of BB bags and picked one that had about the heft of one hundred fifty grams. I tied a 37.5 inch tether on the bag and began throwing. I thought I would have to adjust the mass some but to my surprise the wheel stopped quite nicely with slight backward motion. I took the bag to the scales and guess what: 150 grams.

Then I got to thinking: If the 56 inch 98 gram tether stops the wheel from either wheel diameter then the 37.5 inch 150 gram tether should stop the wheel from the 18 inch diameter. I threw from the 18 inch diameter and again the wheel stopped nicely with slight backward motion.

Quite obviously the wheel is moving 18/12 = 1.5 times faster at 18 inch than at the 12 inch diameter. And 150 is 1.5 time more massive than 98.2. So the wheel is throwing the same because the output momentum is the same 12 * 150 = 18 *100. This then is yet another 'experimental proof ' that wheels throw momentum. The energy of the 100 grams is however 1.5 times greater than the 150 grams.

The thrower has greater respect for the 100 grams than the 150, even this small increase is noticed when throwing.

[pequaide]

One clarification; the difference in the energy is only 50% between the 100 gram and 150 gram missiles, but the overall energy change between the wheel and the missiles is huge.

The wheel is only gently spun but the missiles (BB bags) will fly to the end of the lab.

[Tarsier79]

Although I don't really want to encourage you, an author named William Kenrick designed a "rotator" after studying Bessler in the mid 18th century. From his writings, it appears he was also a momentumist.

http://en.wikipedia.org/wiki/William_Kenrick_%28writer%29 disgorges similar theories on transference of momentum between a large and small mass(using "sliders" in his 2 documents on perpetual motion: "An account of the Automaton...including 2 letters, one from Gravesande to Newton......" and "A Lecture on the Perpetual Motion"

Apparently, he built a "working" model, but later disassembled/destroyed it, after it didn't produce enough power to be a prime mover to drive the machinery of the day. He planned to perfect a more powerful model that unfortunately (but not so surprisingly,) never came into existence.

[jim_mich]

His 109 page book on perpetual motion, written in old English, is readable on Google books. (Here is the link.) A quick skim does not show any drawings. But there are a couple pages that seem to be missing at the very end of the book.

[pequaide]

I am now throwing with 380 grams of drive on the 12 inch diameter while throwing a 98 gram missile from the 18 inch diameter. The tether is a full wrap of the circumference. The missile lifts above the machine as the tether unwinds but its maximum velocity is down and into the floor. This downward direction is determined by where the missile first lifts from the surface of the 18 inch wheel. To change the direction I will have to use a mechanical release. But I will not be able to do that until the throws can be done outside. Down is the best way to throw in the lab anyway.

The velocities into the floor are reasonably violent. It is far more violent than the overhead velocity, it violently accelerates into the ground. The wheel does not stop, which means there is even more motion left in the wheel. Possibly the missile thumps the ground before it secures all the motion of the wheel. I am reasonably certain that this is a decent throw but in the wrong direction. And it is done with little mass difference; 380 at 12 and 98 at 18.

I also used the double wheel as an Atwood's again. I put 1.262 kilograms on both sides at the, little over, 12 inches diameter. Then I balanced one side and then the other side with masses at the, just under, 18 inch diameter. This gave me two masses of .88 kilograms on each side at the larger diameter. Both of the (2 * 1.262 at 12 and .88 * 2 at 18) arrangements would accelerate through half a rotation in 3.72 seconds. The drive mass in both arrangements was only 68.3 grams at the 12 inch diameter. This is a lot of motion for only 68.3 grams dropped .48 meters.

This; only 68.3 grams dropped .48 meters, inspired me to try a low mass difference, slow throw.

You need some speed to make a vertical throw because the missile simply falls off the wheel before the motion of the wheel can catch it. Then the missile is dangling as a pendulum when the wheel tries to pick it back up, I just does not work.

But the 68.3 grams gives you .25 m/sec at 12 inches, and it seems like that is getting close to throw-able. Anyway that is why I tried the 380 g / 98 g throw.

[pequaide]

I cut two 19 inch circles out of ¾ inch plywood and glued them together side by side. I mounted this disk on a .75 inch shaft and placed the shaft into two industrial pillow blocks. When thrown: a full circumference wrap of a tether with 100 grams on the end will stop and reverse the wheel before the tether comes off. That places the rotational inertia of the wheel around 2.5 kilograms. A 280 gram missile can stop it with less than a third wrap of the circumference. The 280 gram throw is an under wheel throw with very little arch, before it slams into the end of the lab.

I suspended 20.7 kg from the shaft and balanced it with a mass placed on the 19 inch circumference. The mass needed to balance the 20.7 kg was 862.8 grams.

I then added 79.8 grams to the 862.8 grams on the circumference. I allowed the 862.8 + 79.8 to drop a certain distance and I timed several drops. It took about 3.3 seconds to cover the distance.

I replaced the 20.7 kilograms at the .75 inch shaft with a balancing mass on that side of the 19 inch circumference. This gives us two 862.8 gram masses. I added the same 79.8 grams to the same side and the drop covered the same distance in about the same time 3.3 second.

Conclusion: The extra force given by the 79.8 grams can accelerate 862.8 grams to .3636 m/sec just as easily as it can accelerate 20,700 grams to .015 m/sec.

Observation: 1/2mv² ½ * .8628 kg *.3636 m/sec *.3636 m/sec = .057 joules: ½ * 20.7 * .015 * .015 = .002375 joules .057 / .002375 = 24

The energy in the wheel itself is the same in both arrangements. The wheel's inertia has been reduced from the 48 kilograms in 48blue to about 2.5 kg , but the results are the same. And the energy of the added masses is a greater portion of the total energy. This is a proof that the inertia of the wheel is of no consequence. It is all F = ma, and the Law of Conservation of Energy is false.

[nicbordeaux]

Peq, you said some posts back that you were throwing at the floor. If your results are as good as they sound, shoot a bouncy ball to the floor, it'll bounce enough. Then all you need to do is a little math, and collect a Nobel prize ,-)

[murilo]

Nic,
I just could note a slightly sign of sarcasm in your above msg.
Am I right?
Best!
M

[pequaide]

There is far more here than a Nobel Prize. But that comes too.

Directional changes are free; you don't need to pay for a bounce. I was throwing down because the ceiling can be destroyed. You can throw any direction you what. If you think these machines can't throw look up 'smokin lamas'.

By making the wheel lighter the difference between 41.4 (20.7 *2) kilograms moving .01515 m/sec and 1.7256 kg (.8628 *2 ) moving .3636 m/sec is now 38% of the output energy. The Law of Conservation of Energy does not work.

The momentum on the other hand remains constant, and the F = ma relationship holds true.

[FunWithGravity2]

close the loop, close the loop, close the loop, close the loop,close the loop, close the loop, close the loop, close the loop,close the loop, close the loop, close the loop, close the loop,close the loop, close the loop, close the loop, close the loop,close the loop, close the loop, close the loop, close the loop,close the loop, close the loop, close the loop, close the loop,close the loop, close the loop, close the loop, close the loop,close the loop, close the loop, close the loop, close the loop,close the loop, close the loop, close the loop, close the loop,close the loop, close the loop, close the loop, close the loop,close the loop, close the loop, close the loop, close the loop,close the loop, close the loop, close the loop, close the loop,

_________________
Si mobile in circumferentia circuli feratur ea celeritate, quam acquirit cadendo ex
altitudine, quae sit quartae parti diameter aequalis ; habebit vim centrifugam suae
gravitati aequalem.

[pequaide]

That is easy. That is easy. That is easy. That is easy.

In the present model the drive mass drop only 6 cm in each rotation of the wheel (.75 inch * 2.54 cm/inch * 3.14159 = 5.98 cm. The wheel will rotate another 1/3 rotation before stopping, for 2 more cm, for a total of 8 cm per throw.

I have used other wheels to throw 15 meters up, so I know that that is easy So I could put a chain of 188 kilograms (1504 cm / 8 cm) on the shaft. And you could throw one kilogram on every 8 cm of rotation. So you have a 15.04 meter chain draped over the .75 inch shaft that has 188 kilograms; one kilogram spaced at every 8 cm apart. The returning side of the chain has no added mass.

With the chain added; the .75 inch shaft and 19 inch wheel becomes an Atwood's. So the Atwood's acceleration would be 188 kg / 88.66 kg + 188 kg + 40 kg ( 88.66 =19 in./.75in. * (1 kilogram + 2.5 kilograms)) (its own mass 188; plus the equivalent wheel mass at the shaft 63.3 kg; and the equivalent thrown mass at the shaft 25.33 kg; and the mass of the chain 40 kg), 188 /316.66 * 9.81 = 5.82 m/sec/sec at the shaft.

So after a drop of 6 cm the chain and shaft are moving: The square root of (.06 m * 2 * 5.82 m/sec/sec) = .836 m/sec. The surface of the wheel is moving 19/.75 times faster = 25.33 * .836 m/sec = 21.18 m/sec; and then we throw.

The one kilogram is thrown above the top of the chain and the loop is closed.