12 times more output than input, dual mechanical oscillation system !

[CuriousChris]

Gravity is not even all around the earth, but it is fairly consistent. The earth is not a perfect sphere. There is more mass around the equator countering the effect of the earths spin. The equatorial bulge. The poles are slightly flattened.

The earth is not solid, the majority of it is a molten rock and metal, the solid part we stand on is actually quite thin. So as the earth spins it deforms and the heavier material tends to move to the equator until its all in a perfect balance.  If it didn't do that it would have spun itself to pieces not long after it formed, and we wouldnt be here to discuss the point.

Quite logical really.


CC

Edit:
Here is a wikipedia article
http://en.wikipedia.org/wiki/Gravity_of_Earth

I know of a better discussion but cant seem to find it. I'll post it if I do find it

[Cloxxki]

Gravity is not even all around the earth, but it is fairly consistent. The earth is not a perfect sphere. There is more mass around the equator countering the effect of the earths spin. The equatorial bulge. The poles are slightly flattened.

The earth is not solid, the majority of it is a molten rock and metal, the solid part we stand on is actually quite thin. So as the earth spins it deforms and the heavier material tends to move to the equator until its all in a perfect balance.  If it didn't do that it would have spun itself to pieces not long after it formed, and we wouldnt be here to discuss the point.

Quite logical really.


CC

Edit:
Here is a wikipedia article
http://en.wikipedia.org/wiki/Gravity_of_Earth

I know of a better discussion but cant seem to find it. I'll post it if I do find it

Thanks Chris.

But can that account for the CF difference between zero at the poles and my fuzzy math estimate of 35% reduction in net weight at the equator? At 1000mph, taking a 6000km radius turn is quite a bit.
I'd love to be violently corrected on my 35% figure, just to know what figures we're looking at.

[Cloxxki]

I have a proposed simplified 2SO

I'll save you my fuzzy math

Specs:
Pendulum 1m, 20kg
Counterweight 100kg
cross beam 3x longer on pendulum side
Starting position pendulum 60º = 0.5m up
I am using g=10 for simplicity
I am presuming a linkage pendulum pivot which always directs centrifugal force directly vertically to crossbar, which itself is presumed level.

Still pendulum: 20kg pulling on string, amplified 3x on CW side reduces CW's load on platform from 100kg to 40kg

A top bottom (static pivot implied), pendulum reaches 3m/s for a centrifigal pull of 180N.
This amplified 3x pulls up on CW for an extra 540N
CW was 400N, and netts -140N (accelerated upwards by 140N force, gravity disabled)

As long a CF from pendulum > 133N, CW is being lifted. this will start before top bottom (t1), and end after it (t2).
Even after the forces equal, the CW will continue to climb for a moments due to it's vertical velocity still present at t2.

The pendulum's pivot, from lifting the CW, loses roughly 3x the altitude the CW gained.

All agree thusfar?

I think dimensions used are very reasonable for building and filming, and weights can be reduced for convenience as mass is irrelevant, only proportions of weights and lengths. Crossbar can also be any length convenient, as long as pivot positioning is proportionate.

[johnny874]

@Jim
Glad to hear you did get your money's worth at engineering school. At times engineers can be so much into the tech, that language gets disconnected. I usually understand technical English decently, but yor sentence building just seems incompatible with my reading ability.

  English is my second language. Americans don't like it either.

[CuriousChris]

Thanks Chris.

But can that account for the CF difference between zero at the poles and my fuzzy math estimate of 35% reduction in net weight at the equator? At 1000mph, taking a 6000km radius turn is quite a bit.
I'd love to be violently corrected on my 35% figure, just to know what figures we're looking at.

Not sure where you get your 35% or what you have based it on.

Here are a couple of more links. the first is plain english with some math, the second is typical wikipedia

http://curious.astro.cornell.edu/question.php?number=310
http://en.wikipedia.org/wiki/Equatorial_bulge