Terragravitic Induction

[Senax]

In the animated .gif below you will see two masses dropping down a slope.
The one dropping down the curve slope arrives before the one dropping down the straight slope. This is because the curved cycloid slope, a brichistochrone, induces more horizontal reaction energy than the straight slope.

https://drive.google.com/uc?export=view ... Ywo5x0p5Le

This energy is 3rd derivative Precession Energy since it is generated at right angles
to 2nd derivative gravitional energy.

It is the enigmatic energy that puts in an appearence (inter alia) in the Faster than the Wind demonstration.

[WaltzCee]

Is Terragravitic Induction ah, well, exactly what is it?

We need a grim thesaurus.

[agor95]

Precession
2nd derivative gravitional energy.
enigmatic energy

I do like using items from their original source

https://en.wikipedia.org/wiki/Brachisto ... chrone.gif

Also a better presentation

https://www.instructables.com/The-Brach ... one-Curve/

It is my choice, but I like to use precession with regards gyroscopes.

The Brachistochrone Curve is a little to linear for that.

Gravity is supposed to be a result derived from a 4 dimensional structure.

How I like to think all energy is enigmatic.

When you are ready to really get into precession let me know Mona Lisa.

Regards

[eccentrically1]

I guess the interesting thing about it is that a point on a circle traces the curve so it's easy to draw or trace it on something and then build it.
It would be interesting to know what approaches the 5 guys took to solve the problem back in the day.
Now if there were only a fast path up the slope :/

[agor95]

I guess the interesting thing about it is that a point on a circle traces the curve so it's easy to draw or trace it on something and then build it.
It would be interesting to know what approaches the 5 guys took to solve the problem back in the day.
Now if there were only a fast path up the slope :/

I suppose there is if you rotate the slope.

Regards

[Fletcher]

For those new to the Brachistochrone Problem it has been discussed here and elsewhere for years.

It is not much of a mystery and is logical from the experimental evidence, but the math is daunting for most.

The thing that Frank seems to shy away from in these discussions is this inescapable fact. Gravity is conservative. As Galileo discovered by experimentation in the 1600's any ball or sphere allowed to roll down any shaped incline will have the exact same velocity at the same vertical height lost. Altho the time taken to arrive at that height will vary on shape of incline.

Having said that whilst a near vertical drop will have the greatest vertical acceleration component it then has to change course and travel across to the finish height - this takes time.

Given a frictionless thought experiment THEY ALL ARRIVE WITH THE SAME VELOCITY - just some get there quicker than others .. and THEY ALL HAVE THE SAME KINETIC ENERGY AND POTENTIAL ENERGY.

Thank you Galileo for discerning this ..

[Tarsier79]

Gday Fletcher
The ball arrives at the bottom with the same velocity. If the ball doesn't drop directly down, it has been accelerated sideways by a ramp, which is a result of the vertical acceleration. If we think about an equal and opposite reaction, is it possible for the ramp to move the ball in the same path, but also be moved by the falling ball due to equal horizontal acceleration and reaction.

So the ball in both instances has exactly the same path, so should have the same speed (and KE), but in the second instance, the ramp has moved as well, which requires energy. At the bottom, do we now have an OU? If not, why not?

[Fletcher]

Gday Kaine .. Good experiment.

Gday Fletcher

The ball arrives at the bottom with the same velocity. If the ball doesn't drop directly down, it has been accelerated sideways by a ramp, which is a result of the vertical acceleration. If we think about an equal and opposite reaction, is it possible for the ramp to move the ball in the same path, but also be moved by the falling ball due to equal horizontal acceleration and reaction.

So the ball in both instances has exactly the same path, so should have the same speed (and KE), but in the second instance, the ramp has moved as well, which requires energy. At the bottom, do we now have an OU? If not, why not?

Didn't have enough time this am to make a sim exactly as you describe (sim attached), so did a quick and dirty to show the gist of the argument.

My sim predicts (as I would without doing one) that Conservation of Energy and Momentum Laws hold true.

Note the velocities and Total RKE and Trans KE of each sphere in each presentation. The guts being that each system has gained the same KE as lost by the sphere in GPE (no frictions to speak of). And when the ramp moved away the ball fell steeper but with less velocity so the energy budget balanced, imo.

Why they lose more KE once rolling horizontally is because I set elasticity in the floor to 0.2 otherwise they bounced around to much for my liking.




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[Tarsier79]

I agree with the sim as far as energy goes. From what I have seen I expected that also, but I don't know why that would be the case.

The velocity of the free-fall differs to the one rolling down the ramp, but the total energy is the same because there is no rotational energy in free-fall.

So what is the difference? The different angle of the ramp compared to its fall will mean the ball has a different rotational KE, but I am still at a loss for why its total linear + rot KE is sapped by moving the object in this case.

[Fletcher]

Mate .. gotta go and do other things ..

Here's another quick and dirty - more along the lines of your original proposal.

I'll give it some deeper thought when I get free time .. atm I'd say off the top of my head inertia's (Momentums) come into the game. The ball has to do Work moving the ramp out of the way. Even tho they follow the same trajectory thei velocites are quite different (in sim-world).

RKE depends on the MOI and RPM etc.

The velocity vector is just a linear velocity. If it takes longer to get to the bottom then it must have a lower "linear" velocity vector.

Everything is more pronounced in this animation and sim (attached).

That's why it'd be a good real-world experiment - one up on Galileo lol.




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[Tarsier79]

Could you remove Rot KE from the equation?..... You could put a 1kg weight pinned inside a "0 mass" roller.

I agree, I think a real world will be interesting, and will most likely match the sim. I will see if I can do a mock-up at some stage.....still haven't managed to do a lot of my other tests.

[Fletcher]

Reduced the mass of the old rollers to 1 gram each (0.001 kg - can't have zero mass) and inserted green 1 kg masses pivot pinned to center of roller - they maintain their original orientation.

Now there is no RKE and only Translation KE of green rollers, and velocity vectors etc.

ETA : look at the red triangle ramp - the normal (right angle) force contact is still quite different from the blue contact normal force with roller i.e. more upwards component even tho it follows the same ultimate trajectory as the blue.

Happy squaring it away :7)




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[WaltzCee]

wow, that's some serious analytical trig-eometry, Fletcher.

[Senax]

Gday Fletcher
The ball arrives at the bottom with the same velocity. If the ball doesn't drop directly down, it has been accelerated sideways by a ramp, which is a result of the vertical acceleration. If we think about an equal and opposite reaction, is it possible for the ramp to move the ball in the same path, but also be moved by the falling ball due to equal horizontal acceleration and reaction.

So the ball in both instances has exactly the same path, so should have the same speed (and KE), but in the second instance, the ramp has moved as well, which requires energy. At the bottom, do we now have an OU? If not, why not?

Mv = mV, conservation of momentum

so the system as a whole gains no momentum energy, gains no 1st derivative energy.

But
Mv² < mV²

so the system as a whole has gained second derivative energy
which is the sum of Mv²+mV² and is always positive since

-v * -v = +v²
and
-V * -V = +V²

Quite apart from anything else, common sense suggests that if the car taking the low road gets to Scotland before the car taking the high road the first car must have got extra energy from somewhere. :o)

[WaltzCee]

holy cor·pus·cle, Batman!!

It's the proverbial power stroke gradient principle FCDriver was BALLYHOOING back in the day.