If a large wheel has rotation with a weight mounted at 3 oclock would that weight be divided evenly at 1,2,3,4,5 oclock to achieve the same force?On the other side of the wheel work is being done ie there is a resistance to this rotation ........
Quote from: mapsrg on April 05, 2008, 04:54:18 PM
If a large wheel has rotation with a weight mounted at 3 oclock would that weight be divided evenly at 1,2,3,4,5 oclock to achieve the same force?On the other side of the wheel work is being done ie there is a resistance to this rotation ........
I believe that torque is greatest at 3 o'clock, because there gravity helps you the most, as gravity has the best angle to do work. Hope I am not missing something simple. I am not an expert.
How we understand weight on a wheel is allot more complicated than you know. But yes 3 o'clock is the strongest point if that is the only weight on the wheel. Each time you spread the weight it changes. For instance if you have 2 weights on the wheel across from each other and spin it the weights will end up at like 2 and 8 o'clock positions not the 3 and 9 but if you pull it up to it, it will stay.
I thought 2 o clock wouldt be the most strong?
I have made the following calculations for weight distribution....3oclock 100% ,that is maximum distance from centre of rotation .This means at the 12 oclock position the distance is effectively 0 as a line drawn down is right on the centre of rotation.At 1oclock it is 50% of the distance and at 2oclock @90%.
Quote from: mapsrg on April 19, 2008, 12:39:09 AM
I have made the following calculations for weight distribution....3oclock 100% ,that is maximum distance from centre of rotation .This means at the 12 oclock position the distance is effectively 0 as a line drawn down is right on the centre of rotation.At 1oclock it is 50% of the distance and at 2oclock @90%.
Interesting, also that means there has got to be a lot of work done on the descending side of a wheel, to lift the ascending side of a wheel.
The work on the descending side of the wheel in raising the ascending side is provided by the differential in distance from the centre of rotation between the two sides.Similar to a see-saw with people of the same weight but differing positions from the pivot point.The trick is to attain this change in position continously to cause rotation of a wheel.The rotation is caused by gravity but the postioning of the weights must be accomplished by some sort of guides be it mechanical, magnetic or electromagnetic.Overunity is the goal and it is attainable...
The distance from 7o?clock to 1 o?clock is the same as the diameter.
The way that a weiht is traveling from 1 to 7 is ~1,57 times the diameter.
Perhaps it helps to calculate the reqired force to repoint the weight.
helmut
a weight at 12 oclock has no turning effect as is a weight at 6 oclock.A weight at 3 oclock has maximum effect.......also a weight at 2 and 4 oclock approaches this maximum @90%.
The energy required to travel around the wheel from 12 to 6 (output) and 6 to 12 (input) can be calculated using the circumferencial travel distances as part of the equation.
PE=m*g*d in joules per second
For a wheel of 1metre r the circuference is 6.24 metres .
if we use a mass of 100 grams per weight
if we move the weight out 100mm on one side and in 100mm on the other
for the falling side we would have a new circumference of 1.1*3.14=3.45M
for the rising side we would have a circumference of .9*3.14=2.82M
Therefore
PE=m*g*d
PE=0.1*9.8*3.45
Pe=3.381 joules per sec output
and
PE=0.1*9.8*2.82
PE=2.2636 joules per second input
the difference is 1.1174 joules per second for one weight in a cycle around an unbalanced wheel
If this offset effect was caused by a mounted smaller wheel that is rotated at 6and 12 oclock then these wheels would be 200 diameter and with a 50 mm clearance between them there would be @24 of them.This is 26.81 joules per second.
the energy needed to rotate a 200 diameter small wheel is
PE=0.1*9.8*0.628
Pe=0.6154 joules per second
Therefore we have excess energy of @26 joules per second from this unbalanced wheel..... :)
I don't know how many know what a crossp roduct is but torque is basicly RxF where R is the vector from the midpoint of the circle to the position and F is that gravity vector. This will give you a new vector. The direction of it will be determinded by the righthandrule/corckscrew and the size is the area of the parrallegram formed by these two vectors. This is also why at 3 o'clock you have the biggest torque contribution since that area is then biggest. ILLUSTRATION TIME!
(https://overunityarchives.com/proxy.php?request=http%3A%2F%2Fbroli.dommel.be%2Ftorque.PNG&hash=7364e70c6dcfc15b6f2e30a3ec0779d738a3c0ae)
I hope that's a bit helpfull.
very detailed
http://en.wikipedia.org/wiki/Torque
in the article they state calculations dont relate to torque due to gravity....
I was thinking that rpm would also effect the power as if the wheel spins at 60 rpm (1 rev a second) the power output is similar to that worked out but at an acceleration of 9.8m/s/s revs will increase to say 120 (2 a second) 180 (3 a second)etc...I will have to read up on this aspect.
The example of 24 wheels given above can have its output doubled by using wheel pairs and increased further by having another inner circle of smaller smaller wheels of 100 diameter at a radius sitable for 24 of these.
since gravity is acting at 9.8m/s/s and the total circumference of the offset paths of the smaller wheels is 6.27 m then in one second the wheel would rotate 1.56 times.If this is correct then the output would be also greater proportionally...40.56 joules per second??????