Roll on the 20th June

[noonespecial]

Let's have a look at Archers diagram shall we.

If you squint at Archer's diagram you will see a lever that is 1 meter on the left and 5 meters on the right. This is according to 'his' scale that each division is 1 meter. I get 5 meters on the right although I can't read all the ticks due to the lack of resolution. This represents a mechanical advantage of 5 and not 10 as you imply.

For the weight on the left to move 2 meters vertically, the weight on the right needs to move 7 meters vertically. This is pure geometry, lay it out on paper and you can confirm. Notice that the great mighty Quinn  pulls something strange out his ass, "the grey line of axia".  What ever in gods name does that have to do with anything I have no idea. Then he procedes to arbitrarily extent the grey line from the end of the lever to the vertical scale on the right. The vertical distance between the intsection points of the grey axia and the vertical scale measures 10 meters. how clever. The real distance that the weight on the right moves is only 7 meters vertically and not 10. This can be confirmed with trigonometry or drawn to scale in Audocad or on a piece of paper. 

Now lets see what we really have; on the right we have 1.2kg x 7 meters or 8.4kg.m worth of work. on the left we have 10kg x 2 meters or 20kg.m worth of work. I'm reasonably convinced that 8.4kg.m of work is not going to convert to 20kg.m? We're missing 11.6kg.m

This confirms that Archer is either very ignorant of physics and geometry or is trying to decieve.

I include a diagram (to scale) to illustrate my point.

respectfully,

onesnzeros

This would be easier if we had a better graphic  . Yes, it appears now that you are correct. When it was mentioned at the bottom of the page that the machine would be approximately 30' in size I assumed there would be a 10 meter arm. At 5 meters, unless there is something else going on, he would need a weight greater than the 1.2 kg indicated. He could use 2.5 kg (2.5*5=12.5)and it would work fine but I don't know if this changes the overall dynamic of the rest of movement.

[ramset]

WELL.... since the other boss [Stephan ] showed up  the armchair ingrates that do nothing  but act like children have sat down and shut up [good idea] @ Broli your good people [not that you need to here it from me] this is a time to work towards a solution not to throw stones    Archer  GOOOOOOOOO.... boy     Chet

[hartiberlin]

Well, I have pondered again and again,
but as in my simulation example I used a 4:1 ratio,
the red weight always has to be pulled up 4 times as high as the yellow
weight falls and this is the killing principle, because it does not work.
I did not see it the first time when I looked at the user Broli
pics.

Also with changing pivot points you can not get any energy gain.

Always there is truth in:

m1 x g x delta h1 = m2 x g delta h2

It is always conservation this way and also via
shifting pivot points or lever arm this equatation stands.

The only thing you might get something is,
when you store the down-movement of a weight in a
spring and thus conserve its potential energy inside
a spring elongation or compression by

1. move weight at 3 o?clock more away from the pivot point (axis in a wheel) ( spring is elongated)
2. fix the weight there to the leverarm,so spring can not yet pullback the weight.
3. the wheel thus is unbalanced and begins to rotate clockwise
4. at about 8 o?clock unfix the weight from the leverarm, so spring can pull back the weight.
5. The stored elongated spring energy will pull back the weight into the direction of the axis and again accelerate the wheel
and compresses the spring, as the weight now is closer to the axis than as in the idle state. Compression of the spring
stores again the energy of the weight.
Weight must be fixed then again until about  2 o?clock to restart the cycle.

Wheel must rotate slow, as when speed gets too fast the centrifugal forces will
influence this principle negatively.

Regards, Stefan.

[broli]

@Stefan;

You have to acount for friction losses of the spring. After a while you won't be able to pass the locking mechanism anymore.

[hartiberlin]

@Stefan;

You have to acount for friction losses of the spring. After a while you won't be able to pass the locking mechanism anymore.

Hi Broli,
as the wheel accelerates,
surely you have to somehow put some part of the additional output energy also back into the
spring mechnism to compensate for the spring friction losses..
Not so easy all together with only mechanical parts..