Roll on the 20th June

[Rusty_Springs]

Hi All
All I can say is to me this has been a good thread for OU because we have debated alot of things to do with OU.
We have gone through gravity, magnetic force, heat, leverage, speed, centrifugal force and friction and I think mosts knowledge has improve because of it which can only go we for OU.
Take Care All
Graham

[MrKai]

Hold on a sec. Archer's still showing us a working wheel next week, right?

In theory.

Presently, he is making Sesame Street videos to explain it. If we follow the trend, he will make short videos demonstrating each of his "working principles" then declare that it should be a piece of cake for any sentient being to "wrap it up".

I kid.

Of *course* he will have and demonstrate a working wheel.

-K

[spinner]

Hold on a sec. Archer's still showing us a working wheel next week, right?

Yeah, sure... The day and the month are fixed, only the year may change slightly...
As soon as he grasps a lever, he'll continue working on his wheel. We've already seen that he finished the hub, axle and support...

If you have a petrol generator handy, you can connect to grid in no time....

Btw, if he doesn't succede, blame Newtonians...

[Bubba1]

Has anybody understood,
how the Egyptian Fulcrum is woprking from the
new descriptions on Archer?s page at:...

Uhh, no.  Don't get it. Of course, I've only read it 3 or 4 times.  If someone took an MRI of my brain right now, it would look like a plate of spaghetti.

From one paragraph on his website:
"So we subtract the weight of the 1.2 metre section from the 6 metres section as already against the opposing end, leaving us with 4.8 metres at 4.02 kilos per metre = 19.296 Kilos to pull down the beam short section to the ground or start point."

He loses me right there.  If you cancel out the two equal sections, there IS 4.8m of long beam @ 4.02kg = 19.292kg, but this is acting @ 3.6m from the pivot, giving 69.46... (kg-M) of torque.  To balance this he would need 19.296kg at 3.6m, but since the short end is only 1.2m long, he needs 57.888kg @ 1.2m to pull down the beam short section, or more closer to the pivot (numbers used here have roundoff errors).

Edit: changed 69.46... N-M to kg-M of quasi-pseudo torque (without g)

[spinner]

Uhh, no.  Don't get it. Of course, I've only read it 3 or 4 times.  If someone took an MRI of my brain right now, it would look like a plate of spaghetti.

From one paragraph on his website:
"So we subtract the weight of the 1.2 metre section from the 6 metres section as already against the opposing end, leaving us with 4.8 metres at 4.02 kilos per metre = 19.296 Kilos to pull down the beam short section to the ground or start point."

He loses me right there.  If you cancel out the two equal sections, there IS 4.8m of long beam @ 4.02kg = 19.292kg, but this is acting @ 3.6m from the pivot, giving 69.46... (N-M) of torque.  To balance this he would need 19.296kg at 3.6m, but since the short end is only 1.2m long, he needs 57.888kg @ 1.2m to pull down the beam short section, or more closer to the pivot (numbers used here have roundoff errors).

Lol, Bubba1, better not... People here are already confused enough! Let's leave the lever thingee...

After all, IF he would ballance the lever BEFORE he attached both weights, WE WOULD all know that "he never lifts a 20kg with 1kg on a 5:1 lever"....
Cheers!