Roll on the 20th June

[MrKai]

Has anyone read Archer's gpage.html? http://www.surphzup.com/gpage.html.

Halfway down this page, it appears that he is attempting to again describe the reset mechanics/methodology/mechanism, but this time, as a weird parable.

I'm sure the dudes that are hangin' in his inner circle know, or think they know, or have been told what he's talking about...and from what I am reading I believe he (archer) or someone else drew some pictures wrt this thing.

What is he saying? It seems like it is "backwards" for some reason. If my...understanding of this is somewhat...correct...I believe he is trying to say that the weight is water, and is somehow removed and replenished based on which end of the thing is up (or down) thus perpetuating the cycle.

Does anyone follow this? I mean it seems like it could work in...theory...but...I'm not seeing the weight distro trick in this setup.

Oh and before anyone says "drinking bird" look it up; it works on a different principle (heat exchange/loss).

[Rusty_Springs]

Doc,

I'm still doing some experiments to see if there is anything unusual happening with Archer's lever. So far I have not found anything, but that's okay - doesn't mean that Archer didn't build anything special just because I can't.

Anyway, I did "learn" a few things along the way (or rather: realize, because the things I have *learned* a looooong time ago seem to predict those things) - not only with levers.

However, my last experiment was inconclusive because I did not have enough material, so maybe you can tell me what the theoretical outcome is. I have a beam. I drill a hole in that beam, exactly at the center. I put an axle thru - this is the pivot. If I now mount equal weights at the end (I do not put them ON the beam, I mount them so that they are centered - does that make any sense?), the lever will "float" wherever I turn it, right?

No what happens if I drill holes thru the ends of the beam (again, centered - so they are on the same line as the pivot), put an axle thru at each end and have weights hanging from those axles - will the beam still "float" (provided that those weights are equal on each side)? Will it make any difference if the weight on one side hangs down 10cm and the weight on the other side hangs down 20cm?

Regards,

Rainer

Hi Rainer
I could be wrong not knowing about levers and maths but I would think the level will drop were the 20cm, the reason I think this may happen is because of the extra weight from the rope holding the weight at 20cm overbalances that side.
Take Care Rainer
Graham

[rainj1]

See the wheel at this video:

http://www.youtube.com/user/redriderno22

Now make the magnetic ramps much closer to the center of the disk (and also the movable rod with magnets at end much shorter) and extend the weights with their rods much further out from the disk.
Now we have the system, where the magnets are able to lift the weights from 7 to 1 o'clock, and the torque amplification factor of the wheel (weights far out against the magnet-lifters very close to the center - a normal lever amplification principle) ensures that the magnetic repulsion wall will be easily overcome.

Basically when you think about the torque at 9-3-o'clock line - it depends on the mass defference at the right and left side of the wheel. So for overcoming the magnetic repulsion (or the lifting force needed) can be calculated from the equasion:

(m1-m2)*(right_hand_long_mass_lever_length) > (m1+m2)*(distance_from_disk_center_to_magnets)

If this equasion is always satisfied, then the disk should always move in righthand CW direction.

Hopefully I'm right. Or close. Any ideas welcome.

Regards,
Rain.

[rainj1]

Oops,

the equasion should be:

(mass_right*rod_len_right-mass_left*rod_len_left) > (mass_left+mass_right)*(radius_from_center_to_magnets)

Or something like that...

--Rain.

[DarkStar_DS9]

I could be wrong not knowing about levers and maths but I would think the level will drop were the 20cm, the reason I think this may happen is because of the extra weight from the rope holding the weight at 20cm overbalances that side.

I guess you're right, but that was not the point I wanted to make - so let's say that I compensate for the weight of the rope on the other side.

Or let's not use weights hanging from a rope, let's use some bars 200mm x 20mm x 5mm, drill holes into them - for one side, we will drill a hole 20mm from the top, for the other side let's use 30mm from the top - and put that on the axles. I should add that those axles are parallel to the pivot.

Now does that make any difference? I guess I'm asking if the CG will change because one weight is mounted higher than the other.

Regards,

Rainer

P.S.: I'd love to illustrate this with a simple drawing (still having problems to express myself in english when it comes to mechanics) - if someone has a link to a free prog for Mac OS X, please let me know