parrallelogram oscillation lever OU?

[mr_bojangles]

this idea stemmed from a previous attempt to use two pendulums, here
http://www.overunity.com/index.php?topic=8594.0

this version uses gears as opposed to a pendulum, and provides a mechanism to reset itself

the attempt is to have a lever that can only oscillate between 45*

ok, assuming all axles are locked

starting with current set up,  we release main wheel, leaving the middle yellow and orange axles locked

this set up of the same sized gears with the  middle locked gives the gear a motion that keeps it upright respect to the event horizon

this is key because it enables us to recoup the angular distance that would normally be lost had i not done this

essentially, every angle the lever moves, would be the same angular distance that we would lose if the counterweights stayed parallel to the lever itself

since we need the weights to move 90*, we make the secondary wheel half the size of the main wheel

this means we only need our lever to move 45* to get us 90* of rotation from the counterweights

my drawing is not to scale and is way off of the ratio of what is needed to work

however, the weight of both upright counterweights cancel out, leaving us with the two red counterweights


in a real set up, the length of each counterweight would be a lot taller, so we could have the counterweight on the right directly over the axle at least, making the resistance on one side 0 units

we now have our weight x distance away falling 45*, obviously the longer the lever, the further distance and more efficient it would be

so we now have a force of (x amount of weight)(45*)(d, distance traveled) acting upon the main wheel

this in turn spins the secondary 90*, reseting the device to go again


so how much force will it take to make these counterweights move 90*?

they balance each other out, the placement of the counterweights would suggest it would be considered the weight of two counterweights lifting the same amount of mass the same distance, making the input seem to be around the amount of friction on the mechanism

after reviewing my drawing i figured out a couple things that are not necessary and simpler ways of accomplishing the same thing, will come later


let me know what you think

[mr_bojangles]

ok, so here are some quick calculations, to make it easier ill throw in some numbers for an imaginary running one

assuming the length of the lever is 4', and a rod extends from each end of the parallelogram upon which two counterweights are placed, 90* apart, as shown in the picture

the counterweight has a 4' rod in between itself and its axle (distance between red weights and grey wheel), meaning the height and length of the set of counterweights is 4'

each counterweight is 2 lbs

now with our current set up, our counterweight on the furthest left, is now 4' (length of rod) + the remainder of the lever, being 2', because the upright support of the counterweights is 2' away from the axle
giving us a total of a 2 lbs force, and because we are limiting the oscillation of the lever to 45*, we now know we have a force of (2 lbs)(a distance of 6')

seeing as how the upright weights cancel out, we can conclude this is our output (assuming no friction)

now, if we let the counterweights fall to their own accord by unlocking the axle, they fall and rest at 45* in respect to the upright support

seeing as how we have two, the total mass is 8lbs, now we need to move it the additional 45*, so it can reset itself and get back to starting position

input is now 8 lbs  @ 45*

the same as, 4 lbs  @ 90* right?

so on a lever, it can lift a 2 lbs weight 45* , because it would be twice the distance away from the axle, and travel twice the distance

for it to be a self runner, the output would need to be able to lift a 2 lbs weight 45* respectfully,



remember now that the output was a 2 lbs weight 6' away from the axle, again limiting the oscillation to 45*, 

2 lbs of weight moving 45*, (output), at twice the distance (4'), can lift 4 lbs 90* half the distance (2') (input)

the output is actually already greater than the input, because it would actually be 2 lbs at 6' away, making the output greater

now after all that, dont forget to factor in the other counterweight, which is now located 2' away from the axle on the heavy side


shouldn't this be OU??

[mr_bojangles]

quick  of the series, this set shows the counterweights underneath, but placement doesn't matter





edited::::: to add pic