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Gravity powered devices => Gravity powered devices => Topic started by: gmbajszar on May 11, 2014, 04:39:26 PM
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(see picture below) If a weight lifter can hold 10 weights vertically, how many weights can he hold on a 45 degree slope?
Twice as many weights.
If we place twice as many weights on a 45 degree slope, it runs lower than the length of the ten weights.
If we connect the triangle with a horizontal line on the bottom, the left side will be heavier!!!
Note: The pebbles have to run close to the bottom of the triangle using a rail system.
George
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In my opinion: The triangle is special. It is the symbol of alien powers.
We kind of sensed earlier in life that the triangle has something to do with aliens.
If we can make energy, then running the triangle with energy in space possibly moves a spaceship since we tap onto gravity/inertia.
It is the fyoochr. It is where we turn into aliens, and our brains will begin to shrink.
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Awww....
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What the math reveals about this puzzle is shocking, simple, unbelievable. A math that a high school first grader can solve.
A very simple puzzle, and a very real one, no catch.
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This is the greatest mathematical puzzle in the history of math.
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I rotated you picture clockwise, and I saw your little man on the bottom of the page with those weights hitting him on the head. LOL
No it is not the greatest mathematical puzzle in the history of math. Add rotation dynamics and it still falls short.
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It is the greatest puzzle in the history of mathematics, because:
a) He holds ten weights vertically. On a 45 degree slope he can hold 20 weights, so the two sides balance out.
b) We measure the length of 10 weights, and the length of 20 weights on a 45 degree slope. 20 weights extend longer down on the right side.
c) We connect the triangle on the bottom, we see that the left side can be proven to be heavier, since we can only fit 14.4 weights on the right side if we can fit 10 weights on the left side of the triangle. When balance needs 20 weights on the right side.
George
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So George, it should be pretty easy for you to make a model using a lubricated slope & continuous chain, & some pulleys to reduce frictions - it should revolve CCW continuously if you are correct - if it doesn't then "Houston, we have a problem" & your math may be incomplete.
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It won't work, but what is interesting is to discover the paradox in the math of this puzzle.
The puzzle doesn't require a genius to understand in what it discloses.
My theory is that it doesn't work for 'other' reasons that man is not supposed to know.
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It is the greatest puzzle in the history of mathematics, because:
a) He holds ten weights vertically. On a 45 degree slope he can hold 20 weights, so the two sides balance out.
b) We measure the length of 10 weights, and the length of 20 weights on a 45 degree slope. 20 weights extend longer down on the right side.
c) We connect the triangle on the bottom, we see that the left side can be proven to be heavier, since we can only fit 14.4 weights on the right side if we can fit 10 weights on the left side of the triangle. When balance needs 20 weights on the right side.
George
You forgot to take into account that the 10 weight's on the left side,travel a shorter distance than the 14.4 weight's on the right side(45* slope) So although less energy is required to move the weight's up the 45* slope,the energy must be applied over a longer distance than that of the 10 weights on the left side. It would take exactly 14.4 weight's to fall on the left side (vertical drop) to pull the 14.4 weights up the 45* slope.
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OK.
You say yo idea dont work.
I'm to dumb to to figure out why it should work, or, should not work.
Spain why it does , or, dont , work.
CANGAS35
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a^2 + b^2 = c^2
If we can fit 10 weights on the left, we can fit 14.14 weights on the right on a 45 degree slope.
What balances is 10 weights vertically, or 20 weights on a 45 degree slope.
(Yet we only fit 14.14 weights on the right side, not 20.)
It was shown that British were extremely bad in math internationally, even math teachers couldn't pass an international math test when all Russian math teachers scored 97 percent on it. I am not implying that there is a UK/US imperialism disorder so people can't figure out high school first grade level math.
How many weights do we need on the 45 degree slope to balance out 10 weights?
Ok, 20. But 20 weights extend lower than the height of 10 weights.
And still, this machine does not work, why. The answer may be Ph.D math level.
George
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George,
The math behind this is trigonometry. It is really needed to understand a lot of physics problems.
The force to move an object up an incline plane is equal to its Mass (m), gravity (g), and the sine of the angle (theta).
f=m*g*sin(theta)
[For ease in demonstrating the math, we need to agree on the mass of the beads and their distance apart. I will say that the beads weigh 1 Lb each, and are 1 meter apart. This makes the math a bit easier to lay out, although we all know that a real beaded necklace would be lighter and the beads closer together. Also I will be ignoring friction to make things simpler.]
On the vertical side of the triangle, two beads weighing 1 Lb each, the Force in SI units is 8.8964432 Newtons
F = 8.896 Newtons (Vertical)
If the height of the 45 degree triangle is 2 (meters), the length of the slope is 2 * sqrt(2) (exactly)... or approx 2.828 (meters). [I use meters, the SI unit, to prevent goof ups]
Now to calculate the force needed to move beads up the incline plane of 45 degrees. There are 2.828 beads on the incline plane (the third bead isn't all the way on the incline plane). The force needed to move the beads is:
2.828 Lb = (1.283 Kg * 9.81 m/s/s) = 12.5795706848 Newtons
F = m*g*sin(theta)
F = 12.580 Newtons * Sin(45 Degrees) = 8.895 Newtons (up 45 Degrees)
Here we show that the force need to move the mass up the 45 degree incline is approximately equal to the force need to lift the mass vertically.
Quick recap....
Each bead is about 1 Lb, 1 meter apart. (1 Lb = 4.4482216 Newtons = 0.453437472 Kg * 9.81 m/s/s)
Calculate lift force. (If you know the weight in Lbs, just convert to Newtons. Stay in SI units, and you'll have no problems.)
Calculate force up incline plane. (you have to calculate the lengths of the triangle using trig if you use anything other than 45 degree triangle with the rule 1 Lb, 1 meter apart)
F = m*g*sin(theta)
Now to see what happens when you go to a 30 degree triangle.
Lift force is the same.
F = 8.896 Newtons (two one Lb beads 1 meter apart = 2 Lb * 4.4482216 = 8.896 Newtons)
Because the vertical length is the same, we need to know the length of the hypotenuse (aka the number of beads on the hypotenuse).
Hypotenuse Length = Vertical Length / Sin (30 Degrees)
H = 2 / Sin(30) = 4
So there are 4 Beads on the hypotenuse. (4 Lb * 4.4482216 = 17.792 Newtons)
The force needed to move the beads up the incline plane of 30 degrees is:
F = m*g* sin(theta) (Note that m*g is weight in Newtons)
F = 17.792 Newtons * sin(30) = 8.896 Newtons (Note that the force needed to move an object up a 30 degree incline plane is 1/2 the vertical force.)
F = 8.896 Newtons
Again we show that the lift force and the force needed to move the mass up the 30 degree incline plane is exactly the same.
No matter what the angle of the incline plane, using evenly distributed masses on a string, the lift force will always equal the force need to move the mass up the incline plane.
I hope this helps...
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What level math is this? Third grade high school?
Interesting that logically the mind assumes that on a 45 degree slope it is twice as easy to pull weight up.
I have a good similar puzzle for you, sir.
It cannot generate free energy, but what I want it to do is to move a spaceship in space.
I attached a picture below. We set up an experiment that begins with being in space in a long rocket. An astronaut throws a heavy weight downward from the top of the rocket, which then makes the rocket move upward until the weight hits the bottom of the rocket when the rocket stops moving. But for a few seconds until the weight flew from the top of the rocket to the bottom, the rocket was in motion upward.
Now, if we analyze the picture below, we use a shotgun to fire two weights downward. As the weights are fired, the rocket moves upward. But then the weights make a 180 turn, which should reverse the momentum of the moving rocket to a full stop. Only when the weights hit the top of the rocket again, I believe we generate motion for the rocket.
Question: Is this a valid theory?
George
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George,
This is first year physics, either high school or college. Normally, a course in trigonometry is required before taking the class. It's the trig part that can trick you up if your not careful. AND, making sure you stay in SI units... it's really easy to mess things up. A good place to learn is Khan Academy. Plenty of video's to get you rolling fast. You don't need to create an account, that is if you don't want progress saved. And, did I mention it's free?
Hope this helps. Good Luck!!!
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Wait, I changed my comment above yours just when you answered. I added another question if you don't mind.
Thanks,
George
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What level math is this? Third grade high school?
Interesting that logically the mind assumes that on a 45 degree slope it is twice as easy to pull weight up.
I have a good similar puzzle for you, sir.
It cannot generate free energy, but what I want it to do is to move a spaceship in space.
I attached a picture below. We set up an experiment that begins with being in space in a long rocket. An astronaut throws a heavy weight downward from the top of the rocket, which then makes the rocket move upward until the weight hits the bottom of the rocket when the rocket stops moving. But for a few seconds until the weight flew from the top of the rocket to the bottom, the rocket was in motion upward.
Now, if we analyze the picture below, we use a shotgun to fire two weights downward. As the weights are fired, the rocket moves upward. But then the weights make a 180 turn, which should reverse the momentum of the moving rocket to a full stop. Only when the weights hit the top of the rocket again, I believe we generate motion for the rocket.
Question: Is this a valid theory?
George
It would be easier to just keep the barrel straight. When you fire a shotgun on a friction-less surface, ice for example, the shot leaves the barrel (quickly) and you move backward. On regular ground, your weight and friction keep you stationary.
With a curved barrel, the shot encounters the barrel wall, imparting forward momentum to the system... AS it's turning around. In other words, your cancelling your backward momentum - up to a point... If you gained forward momentum, the shot would have to be travelling faster down range than the exit velocity from the shell casing.
You only get back what is put into it. That is why it would be easier to use a straight barrel. You would be using the most momentum at the moment the shot was fired. Systems like this are 'expelled mass systems', basically rocketry. With these systems, you have to have a whole trips worth of fuel on board, be it rocket fuel or bullets in this case. This extra mass makes it exceedingly difficult (read expensive) to go into space.
One area being looked at for space travel is a real warp drive. Research Miguel Alcubierre's warp drive, then "warp drive mechanics 101". Basically it states that excluding energy between two plates creates a 'negative' energy density in space. This essentially warps space/time. As the plates get closer together, the Casimir force increases. However, as you get down to smaller scales the surfaces are no longer smooth and the effect cancels out. The math on this is brain busting, and I've yet to grasp the whole thing mathematically.
I'm glad to see you thinking... never give up!!!
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Here is a theory in the picture below that I speculated on 8 years ago. Something that I understand, but I don't think others will.
It supposedly moves a spaceship in space somehow. Thanks for your help,
George