Posted by Joel Lee Lewis (24.197.38.131) on May 14, 2003 at 11:36:22:
In Reply to: Re: Any reason why this idea wouldn't work? posted by Nick Hall on May 14, 2003 at 10:14:08:
Hi Nick. Thanks for the reply.:-)
Yes, I considered that. But it's my understanding that objects always fall(and therefore rise) at the same rate. For instance, if the weight's were cylindrical, and the one on the 'descending' arm was released and allowed to roll up an incline, it's my understanding that it would still reach a height 'allmost' as high as it started from. I'm not concerned with any energy gained from the momentum of the weights themselves, but only with their relative positions and ballance. My base assumption for the entire concept is that whichever side the lever is unballanced toward will descened at the same speed weather there is half the rellative weight, 2/3 the rellative weight, or no weight on the other arm, thun alway giving the weight the same inertia and potential to rise the same distance. If I'm mistaken in that belief, then the whole concept falls apart; if I'm correct, then it 'should' work-any potential energy due to the movement of weithts THEMSELVES is trivial to the concept-it's their rellative position vis-a-vis each other that I'm thinking of.
And forget a wheel. A wheel is, in a sense, nothing more than an 'infinite lever', so to speak.(It is my personal belief that any concept for an 'overballanced' wheel must first be applicable with levers. Indeed, using levers as a starting point may well make conceptualization much simpler and help to expose false leads) I am NOT thinking of something 'raising' the weight-I'm thinking of the weight raising ITSELF, while the lever remains vertical and stationary, due to the momentum gained from it's descent, recontacting the(still vertical)lever closer to the pivot, and thus altering the balance of the lever so that it returns to vertical. As I said, my base assumption is that the initial descent will be at the same rate no matter the 'relative weight' which is being raised on the other side, thus imparting the same momentum to the weight and therefore the same potential to rise. THAT is the vital question to me.
Incidently, I hope people keep the links coming-the one posted by Stefan Hartmann describing how a 'reactionless' propulsion engine could work really gave a great explanation. Hey, but every kid that's ever sat in a grocery cart and started jerking it forward knows it's possible anyways!:-D
: Joel wrote:
: : Okay, here goes: imagine a simple lever with two equal weights, one on each arm, and one twice the distance from the pivot as the other. like this- ____o_____0________o Got it?
: : ^
: : Okay, now imagine letting the lever swing untill it is sticking straight up and down. Picture that moment in your mind.
: Let's put numbers on this Joel. Assume 1 Kg masses. Assume G = 10, each mass therefore weighs 10 newtons. Assume the whole thing turns very slowly clockwise, and as it turns we 'tap' the energy out into some kind of storage device.
: The critical question is: how much energy will we get?
: Assume the weight at the shorter end (left) is 1 metre from the pivot. The one at the longer (right) end is ==> 2 metres.
: Total energy given out when it rotates clockwise to vertical position is loss of PE of right minus gain of PE of left.
: Left gains 1 metre. PE = mgh (assume G=10) = 1*10*1 = 10 joules.
: Right loses 2 metres. PE = mgh = 1*10*2 = 20 joules.
: ==> total theoretical energy output (assuming no losses) = 10 joules.
: Now, the energy needed to raise the lower weight to the 'same distance from the pivot' as the top weight = m*g*height rise = 1* 10 * 1 metre ==> 10 joules.
: Sadly therefore, Energy output = energy needed to move lower weight to balance point with upper weight with respect to pivot.
: Actually it is slightly 'worse' than that because the true starting point would mean rotating the - now balanced - wheel anti-clockwise to a point with a very slight slope to the right, allowing the right hand weight to roll to the right (2 metres from pivot), then lifting it back a bit more t