BesslerWheel.com

Friends,
If you place a fast rotating wheel on the platform, it will lock this platform against the absolute system of coordinates, like a star.
Then the Earth will rotate around the platform, but we will think that the platform rotates.
I suspect this is what Bessler used in his wheel.

Found this patent, does it actually make sense? https://patentscope.wipo.int/search/en/detail.jsf?docId=WO2006001686 (EN)The invention relates to a torque multiplier consisting of two parts which take the form of crosses comprising four arms disposed at 90° in relation to one another and which are ...

In a Newton cradle, all the weights are attached by the string, which means that they are able to make circular motion, or at least an arch. It mainly applies to the first weight and the last weight. At the same time, Newton is using a formula of mv for the conservation of the momentum, not mvr, at ...

Is it? It is the most generic thing there is. Whenever a force is applied (F=m*a) over some distance (s) we have spend some energy (E = F*s = m*a*s) As said, it's a mathematical equality for whichever kind of distance or acceleration (could be centripetal for instance). Except it doesn't work when ...

We can determine [t] to be √(2·h/a) The velocity after dropping height [h] will be: v=a·t = √(2·h·a) When v= √(2·h·a), then v²=2·h·a, or ½·v² = a·h This is all mathematics, in this case derivatives of kinematics: [h] is position, [v] is change of position ...

The ballistic pendulum is a classic example of a dissipative collision in which conservation of momentum can be used for analysis, but conservation of energy during the collision cannot be invoked because the energy goes into inaccessible forms such as internal energy. ???????? http://hyperphysics.p...

Just want to point out that while everybody is saying that conservation of
energy can't be used because of a "heat" losses, apparently there is no
heat losses when they use conservation of the momentum.

Firstly, I didn't bring matches into this discussion, they are irrelevant for the case of ballistic pendulum Secondly, once bullet embeds in the wood, it becomes very well thermally isolated by the wood itself Thirdly, I took the data from the official source, and I believe this type of heat should ...

I took data for lead from the above table
http://www.engineeringtoolbox.com/laten ... -d_96.html

it takes 4 matches for 5000 J

A four-inch wooden kitchen match consumed completely generates approximately 1 BTU (=1055 J)
So it should not only melt the lead, but burn a part of the wood enclosure as well.

Just of the top of my head, 10 gram lead bullet with the speed of 1000 m/sec should have a kinetic energy of 5000 J (mv^2/2) ends up in a pendulum , weight 10 kg raises up 10 cm, total energy = mgh = 10 x 10 x .1 = 10 J To melt lead we need 22 kJ/kg http://www.engineeringtoolbox.com/latent-heat-melt...

I wonder if somebody actually calculted how much heat is being generated
in a ballistic pendulum.
The amount of heat, probably, should burn everything around the test site,
if the conservation of energy actually worked.

So the energy is only created when motion is transferred from the large mass to small! And when motion is transferred the opposite way, the energy is destroyed, like in a recoil, where a large mass of the shooter kills the energy of the kickback of the bullet, like in the ballistic pendulum! This ex...

What's really interesting - when they calculate a ballistic pendulum,
they don't use the conservation of energy equation, but use a conservation of
the momentum instead!

Suppose you have a 19 kilogram rim mounted vertically (horizontal axis) on a quality bearing, and you place a string around the circumference of the rim with a one kilogram mass on the end of the string. F = ma gives you and acceleration rate of 9.81 N / 20 kg = .4905 (or 1/20 * 9.81 m/sec/sec ) = ...