Roll on the 20th June

[Mark69]

Rusty, thats the benefit of a fulcrum.  Even though the weight is the same on both sides, the distance to the balancing point is different.  The side with the longer arm will lift the side of the shorter, even if the weights on both ends are the same.  For each side, multiply the weight (which is the same on both sides) by the Arm (distance from balancing point).  There is where the power will come from.

[Rusty_Springs]

Rusty, thats the benefit of a fulcrum.  Even though the weight is the same on both sides, the distance to the balancing point is different.  The side with the longer arm will lift the side of the shorter, even if the weights on both ends are the same.  For each side, multiply the weight (which is the same on both sides) by the Arm (distance from balancing point).  There is where the power will come from.

Yes mark thats for half a turn and then you have to push the weight back up nulifying any gain. Oh and in the case of the wheel add in two magnetic walls to break through which also takes away most of the gain before you even start the uphill push.
I think people forget to look at the full picture not just half of it.
Take Care Mark
Graham
Oh Try the answer to your magnetic question is they balance ever way as long as nether are closer.

[MrKai]

These are for the wheel, and yes, he sounds like a rational guy, teaching.as opposed to his written rants

There are here: http://www.youtube.com/profile?user=newtonsend269

[Rusty_Springs]

Hi All
Its not hard to prove me wrong just show me a working model.
Oh and I remember Dusty showing a great biuld and I also remember him saying I put the extended arms over the magnets and it seamed to slow it down.
Take Care All
Graham

[purepower]

ATTENTION ALL! Please read. Though it is long, PLEASE DONT SKIM! It concerns everyone and is very important.

Okay, here we go.

So we are absolutely clear on what I am discussing, this entire post is in regards to the effects of extending the weights of on the rods of the wheel on both ends.

First of all, fletcher seems to be the only one that has understood my previous posts one the issue. Please see posts #2236 and #2244.

Okay, everyone seems to be keen on the visualization rather than hard math, so I will try to explain this as best I can for everyone to try to picture what is going on.

Imagine a wheel with a single rod across its center (for simplicity), and the rod is free to slide off-center a distance x to either side of the center of the wheel. Now, the wheel is weightless and so is the rod (again, for simplicity). However, there are two weights located at the ends of the rod. We also have the ability to extend the weights out a distance y from the ends. The length of the rod is 2L, and the weights are of mass m.

Everyone still with me? Good. Now, if we calculate the torque of the rod, this would be equal to:

Moment (or torque/leverage)= m*(L+x) - m*(L-x); which simplifies to = 2mx.

Now if we were to shift the rod over to its other position, the torque would be equal to -2mx. So the magnitude of the torque is the same, but it is now in the other direction.

All together still? Great, lets continue.

Now, lets move just the weights out to the extension point a distance y away from the center. Now if we calculate the torque, we find:

Moment (or torque/leverage)= m*[(L+y)+x] - m*[(L+y)-x] which simplifies to 2mx.

Interesting! So shifting the weights out an equal distance from the center gives us NO ADDITIONAL TORQUE! Again, if we were to shift the rod to its other location, the torque would be equal to -2mx. So the magnitude of the torque is the same, but it is now in the other direction.

Why is this? Well, if we shift one weight a distance y, we gain a torque of m*y. But if we shift the other weight a distance y, we lose torque of m*y, so the two CANCEL EACH OTHER OUT!

Now some are saying to themselves "but Archer had me do a test at home and I saw for myself that this helps break the wall." To that, I say "good observation, but you fail to see why this helps break the wall."

This has absolutely nothing to do with torque. As I just showed you, the torque remains exactly the same. What this does do is change the angular momentum.

Angular momentum= m*r²*w

So the angular momentum increases exponentially as the weights move out. This helps break the wall because it becomes more resistant to the force of the wall. However, this does not fix our wall problem for one simple reason, explained below (and in another post, modified for this discussion). The previous post was for the force of friction, but the same still holds true for magnetic forces (walls).

"What you need to understand is this. Momentum is a state of a body (similar to energy; p=m*v, E=.5*m*v2). Friction (or a magnet wall) is a force that alters the state of a body.

Since friction (or a magnet wall) is a force that will always act against the motion of a body to lower its speed, momentum will always decrease in the presence of friction (or a magnetic force). No matter how fast we spin the wheel or how much it weighs, the friction (or wall) will always win and decrease the momentum. All we are doing by playing with the variables is extending the amount of time it takes for the wheel to stop.

The inevitable truth is wheel will always stop at some point in the presence of friction (or magnetic wall), unless there is another force present to counter the effects of friction."

Tie it all together now. Moving the weights out does not alter the torque, as shown by fletcher. What it does do is increase angular momentum, as seen be those doing Archer's demo. However, increasing angular momentum is not the "quick fix" we are looking for to make the wheel work.

End. Everyone understand? Do I really need to waste time making a video? Before you say yes to that last one, reread the post, think really hard, and if you still dont get it then ask...



Now, for the personal replies:

@Mark69

To purepower:

I am reading one of your above posts and really, have you ever used tools in your life???  Tell me this, if you have a bolt the wont come off with a regular 1/2" rachet, what do you do?  You get the 1/2" breaker bar that has a longer handle to give you more TORQUE on the bolt.  Well guess what, thats exactly what happens here when the rods shift.  You multiply the force of the weight on the end with the longer arm.  If you look up what torque means, it is a measurement of twisting force.  The rods are applying torque to the wheel and the further out it is, the more the weight is multiplied.  When you look at a automotive engine, you really want a larger torque number, because horsepower is not worth crap.  It is the torque that, you know, the crankshaft turning.  Why dont you go ask a race machine shop why it is almost always better to put longer rods in then larger pistons.  It has nothing to do with the weight of the piston either, its about rod ratio. 

You have heard of a "torque wrench" havent you?

Mark

I have addressed this many times. Moving the force out on one side does increase the torque. This is why torque wrenches work. But like Archer's demo on the site, this is only one half of the picture. There is the other side to account for also, which is extended as well, which increases its torque equally and in the opposite direction. This would be like taking your torque wrench and having a buddy work against you with the same mechanical advantage that you just gained. LOOK AT THE WOLE PICTURE, NOT JUST THE PART YOU WANT TO WORK IN YOUR ADVANTAGE.

@LarryC

Forgive me if this was brought up before, not a regular on this thread, but everybody seems to be so concerned about fulcrum math and I'm confused as to why so many ignore the most important of Archer's statements under THE EGYPTIAN FULCRUM section that makes the 5:1 lever work like a 20:1 lever:

If any of those who still choose to ignore these statements would please test the difference between a 1 KG weight resting on your head and 1 KG falling from a meter on to your head. If some sense is not knocked into your head, then there is no way it can be explained.

Regards, Larry

Again, Archurian analysis fails. A weight resting on your head is a statics problem, a weight falling on your head is a dynamics problem. The difference has to do with impact/momentum equations.

STATICS

F=m*g

DYNAMICS

F=(m*V)/t

m-mass of weight
g-9.81
V-velocity at impact
t-time until falling mass comes to rest, usually <.01 seconds

In the lever, this is again something I have addressed MANY TIMES BEFORE. The weight falling builds kinetic energy before it engages the 20 kg mass. This KE converts to PE in the large mass. No magic. Read post #1628, #1647, and #1857 for more detail.

-PurePower