Roll on the 20th June

[ramset]

Spinner thanks for your responce one way or the other I would never take your money  Chet
and yes sceptics    @Dr Stiffler   @Thane Heinz  @ Giantkiller @tesla 2006  etc etc lots of sceptics anytime you go outside the box

[canuck22]

I wish I could draw this, am has the correct idea, there are no springs no magnets, and maybe saying a cam is incorrect

Imagine a fixed piece of metal in a certain shape, as the weight comes around to the 7 oclock position it rides the bar, cam, sheet, call it what you want, it pushes the weight in towards the center

then after the 1 oclock position a bar pushes it out

to me it is a cam

peter

[ramset]

@spinner is there any wall to a lever [such as the wall in magnets] a place that you stop because the effort is less than the work out    ARCHER claims to be able to recycle his lever it seems at 5-1  with 1 unit moving 20 units this is possible ? and how can you say impossible without knowing the complete design

[dirt diggler]

@ Canuck22

I understand what you are try to do, however I believe that this has been tryed many times in the past.
the problem is that the friction is too much, and no matter how much leverage on the one side, it still must drag the weights in, and this prove too much.  that is why the wheel here might hold some promise, because the friction "should" be much less because nothing actually touches to move the weights to the heavy side.
By all means though, give it a try!!  maybe there is some combination of bearings/ materials that is "slippery"enough that it is possible.
I'd probably try it too, but I'm already working on an Archer wheel, and I think I'm going to try the lever, if I can figure out how and what the control rods are.

[spinner]

I have a question

Now please follow me for a bit here. As shown and talked about, a arm on a wheel of equal length from center and with equal weight on each end will find a balance at 9 and 3 oclock.

As I understand it, you need to move the weight at 9 oclock towards the center of the wheel
for the wheel to become unbalance and then to rotate.

Consider that the wheel is moving in a clockwise rotation. I have read every post here and it seems
everyone agrees that the weight loses all kinetic energy at about the 7 oclock position.

I have not posted due to not wanting to feel stupid, and I have tried to draw this, but I have not been
able to figure this drawing program out.

So to the question.

How far to center would you have to move the weight for the wheel to turn, and for the wheel to continue to turn?

My idea is to put a sliding weight on the arm and to use a cam to move and hold the weight at the center when it arrives at the 7 oclock position and then to move it out to the end of the arm when it arrives at the 1 or 2 oclock position.


Any ideas as to if you think this would work?

peter

Now, reread previous posts... Please.

The ideal lever is the one without it's own mass/weight. WEIGHTLESS! AND perfectly straight... Without the pivot center offset... THEORY.. where the lenght of the arms ratio is inversely proportional to the masses... 5:1 lever and 1:5 mass... OK, Archer's Newtonian problem (although the lever math is much older than Newton...Lol...)
In practice, a ballanced beam (if it's asymetrical, it must be ballanced, too!) is similar.
Or a perfectly balanced wheel...

In such situation, perfectly identical weights on a perfectly identical arms (radii from the center) are in balance no matter the angle....

It's logical that a gravity inpact on a mass placed on the circumference of the wheel has maximum effect at 3(or 9) o'clock... And minimum effect at 12 or 6 o'clock... After all, in static situation, one can evaluate gravity effect on a mass/weight on the circumference of a wheel with sin(angle from vertical position... ( 1o'clock (30deg) is 0,5*(mass)  , 2o'clock is 0,866*(mass) and  3o'clock is 1*(mass)...)

Hey, famous Leonardo daVinci made the first analysis of a 'gravity wheel'.... Without trigonometry... His findings are still valid - the gravity impact on a weight fixed at the circumference of the wheel is proportional with the horizontal displacement of the weights from the axle... "Anno Domini  1490".... Imagine that.
More than 500 years ago... And today we're still strugling with the lever mechanics....

"Trading height for width" is an expression which some people here know what it means.... Hey, Fletcher? Rlortie? Hello!

Cheers!