Idea I had since elementary school

[Tarsier79]

Fletcher is right, he is talking about the levers themselves being unbalanced.

Other people (like myself) that are talking about weight position are assuming the levers are balanced.

To illustrate the weight position argument, see the diagram below. The red and blue weight are both 1kg. no matter where you fix the blue weight on the purple line, this wheel will balance, because the horizontal distance from both weights to the axle is equal. If you fix the blue weight above the axle, it will be unstable, and if everything isn't perfect, the wheel will settle at 180 degrees.

Now if you have your weights, and attach them to any arrangement of levers (provided they are balanced to begin with, like the point Fletcher is making) and attach the weights at equal horizontal distance, then both weights will apply equal torque to the levers, regardless of their shape.

Hope this helps.

[Fletcher]

Guys .. I'm out of time & out of town for a few weeks - knock yourselves out.

What I'd suggest is that you reverse the procedure i.e. turn the lever arrangements upside down [hanging on the pivot] so the CoM is BELOW the pivot - that's quite easy to build in your shed - now see where it keels & that's where the CoG/CoM is - flip it vertical again in that same orientation [in your mind] & that's also where it is balanced.

Dan .. IINM what we were investigating was levers with the same radius of operation from the pivot - to make the right angle elements the same length as the hypotenuse would not show that.

Preoccupied .. I don't want to spend any more time on it - eventually you will see that where the system CoM is determines the torque direction - the radius of operation is important if the CoM each side of the pivot is also the same - clearly in your arrangements they are not - leverage is usually described for simplicity reasons as a see-saw lever over a pivot - but it's pretty easy to see that it is the radius of operation that is important ONCE the leverage device is balanced.

[rlortie]

My past explanation of the swastika;

Draw a swastika, inside a circle, place four weights upon it's points. Now erase the swastika, what do you have? Four symmetrical weights in a circle.

No matter how many right angles or crooks and turns you have in a lever, the only thing that counts is the distance from the axis/fulcrum to the weight.

I have seen to many submitted ideas where the designer believes that some compound configuration of a lever is going to help. It does not!

Ralph

[preoccupied]

I've mentioned a couple times that I am probably wrong because people tell me the science states that the radius from the axle determines the distance in the formula Work Done = Force x Distance. I referred to torque in wikipedia and couldn't understand the equations but I got to the part about right hand grip rule and thought it applied to this situation. Does it?

wiki said
"A force applied at a right angle to a lever multiplied by its distance from the lever's fulcrum (the length of the lever arm) is its torque. A force of three newtons applied two metres from the fulcrum, for example, exerts the same torque as a force of one newton applied six metres from the fulcrum. The direction of the torque can be determined by using the right hand grip rule: if the fingers of the right hand curl in the direction of rotation and the thumb points along the axis of rotation, then the thumb also points in the direction of the torque."

http://en.wikipedia.org/wiki/Torque

In the previous animations I believed that the fulcrum on the right animation was the elbow and then also the axle. What is a fulcrum?

wiki refers to this article for fulcrum
http://en.wikipedia.org/wiki/Lever

This article mentions a crow bar and I think that is the type of fulcrum the elbow would be. Who knows about the right hand grip rule? Nothing I've found specifically mentions how the mechanics of a right angle bar works. After I test the torque using a torque wrench I will update this post again and if I'm right I will gloat.

[Fletcher]

What is a fulcrum ?

Fulcrum = Pivot = Center of Rotation [CoR].

i.e. the point of application of the principle of Leverage.

N.B. Leverage is quite different from torque.

A see-saw lever with 1 N force at 6 m horizontal distance = 2 N @ 3 m horizontal distance from the fulcrum - the FORCES are balanced but unless both masses [Weight Force] each side of the fulcrum are the same [e.g. 10 kg each side] then Weight Force [Weight] is not balanced - if it is not Weight & Force balanced then the arrangement will have torque & rotate about the fulcrum to find its lowest position of Pe [keel].

[rlortie]

Having some what of a diversified education and experience in mechanical and electrical, I fee confident that the only right hand grip rule out there refers to magnetic fields in a coil.
http://sciencecity.oupchina.com.hk/npaw ... p_rule.htm

Go here to learn about three classes of levers and fulcrums;
http://www.worsleyschool.net/science/fi ... page2.html

Ralph

[preoccupied]

Basically I think the right angle bar turning into the inside area of it will produce mechanical advantage. In my previous animation it is the picture on the right side. What I think I am being told is that the radius from the axle determines mechanical advantage. If that is true then when a right angle is being lifted it would not produce mechanical resistance. This next animation is hypothetically what could be created if I the opposite of what I believe is true and that is the radius from the axle, despite there being a right angle bar, determines mechanical advantage. When the right angle bar is down a new lever requires greater radius to reach it from the point it would go up when it is lifted and after the weight rolls down the ramp. The extra radius needed to reach the right angle bar allows a large lever to be created that would lift the heavier weight back up as it sets the weight back on the right angle bar. The axle of the long high lever is attached to the axle of the weights pivot so that when the high lever falls the weight pivot is moved back in place. The high lever's axle is a ratchet so that it only responds in one direction and has a counter weight to put it back up into place. This is a gravity driven device.

[preoccupied]

Maybe some distance advantage is gained with the following animation. If only the longest part of the lever is used the distance gained would be relative to the triangles created by connecting the lengths of the squares to the top pivot. Lets say the length of each individual square is 2 and the length from the square to the top pivot is two also. sqrt(40) would be the length from the bottom left corner of the square to the pivot if the square is vertical to the pivot, if the square is level horizontally, because the sides of the triangle are 6 and 2. sqrt(8) is the length of the top left corner of the square to the pivot because the sides of the triangle are 2 and 2. sqrt(40) or 6.32 - sqrt(8) or 2.83 = 3.49. 3.49 should be the minimum distance of the longest lever that can be pushed on the other side of the top pivot from the top part of the square. I don't know if this would create any advantage but it is a new way of looking at turning a gear by using a lever. The idea behind the swastika and the previous animation about shifting the radius is to use levers to create mechanical advantage. The lever might have a mechanical advantage of about 3.49 - 2.83 = .66 but I don't know how to calculate mechanical advantage. If there is a mechanical advantage of .66 then is it enough to turn a larger gear with a farther distance to turn the lever without losing all of the mechanical advantage?

[Tarsier79]

Gday Preoccupied

Have you seen the MT series Bessler produced?

http://www.orffyre.com/mt1-20.html

MT is a series of unworkable designs Bessler had before he supposedly worked out the secret to his "perpetuum mobile." Once you realise why each one doesn't work, then you will understand why mechanical advantage in a gravity field using leverage causes the Centre of mass to fall in a system.

I do like both above diagrams, variable levers are hard to design and use. As far as mechanical advantage goes, a study of simple machines as was suggested before would be greatly helpful to understanding how to achieve this.

Cheers

[preoccupied]

In the previous animation lets say ten newtons are required for shortest part of the longest lever on the other side of the top pivot to push. How much input force would be needed? In this youtube video he says F2*D2=F1*D1. F1=10N, D1=2.83, D2=3.49, and F2 is something. F2*3.29=10N*2.83. F2=8.6N so output force over input force 10n/8.6n = MA = 1.16.

http://www.youtube.com/watch?v=pfzJ-z5Ij48

EDIT
I'm not sure about how gear ratio works but I guess it is directly a ratio. So the ratio between the size of the circles needed would give me the numbers for the gear ratio. The driving circle circumference is 3.14(sqrt(32)=5.65)=17.74. The driven circle circumference is 3.14(sqrt(24.36)=4.93)=15.48. Driven gear over the driver gear = 15.48/17.74=.87. The driving square is larger and turns .87 times for every one turn of the driven gear. I think 1/.87=1.15 is the amount of MA needed to have the input of an identical device be the output of this device. 1.15 is less than 1.16MA so the previous animation should be able to add mechanical advantage to mechanical advantage of about 0.01MA. That's not much though so resistances could cut down that quite a bit but if it's true and MA is achieved enough to be the input into an identical system at a increase then a Perpetual Motion machine can be created by exchanging force between multiple machines using a thrown rock or something else as long as more squares and levers can fit. Although because there is only a difference of .01MA in my math it could be equal due to approximations.

[preoccupied]

I'm going to use F2*D2=F1*D1 for radiusshift11 and lets say the distance of the 2 bars of the right angle bar are 2 ft. The distance from the axle would be sqrt(8). I assume the distance from the axle determines mechanical advantage. Lets say the bar attaching to the big weight length is sqrt(8) or 2.83. Lets say the small weight is 1N and the large weight is 2N. Both the distances are the same because both are 2.83ft from the axle. So F2>F1 to move the weight up the axle and because 2>1 the weight will move up to the top pivot and ramp. The radius of the long lever is sqrt(20) or 4.47. The long lever now has to lift the heavy weight. Distances are the short lever 2.83 and the long lever 4.47. How much force is needed to balance the two levers? F2=2N, D2=2.83, D1=4.47, and F1 is something. F1=.79 so output force over input force 2N/.79N=2.53MA. The small weight must therefore be (small weight) > (heavyweight/2.53) for the large lever to pull the heavy weight back up horizontally. That is more than enough mechanical advantage to do this with the extra resistance and gear ratio increase. Lets look at the gear ratio increase for making the heavy weight go horizontal from the fall of the long lever. Oh my, this would be more complicated. Anybody want to help me with that? The long lever falls < 1/4 rotation and the heavy weight lifts 1/4 rotation. The distance from the lower pivot to the higher pivot and ramp is equal to the length of the right angle bar (2ft). So 4 ft down and 2 ft across from the circles vertical radius is where the long lever stops and just a slight amount less than that in rotation because the lever starts below the ramp. Whoever helps with this last part will be more competent at math than I am. What is the gear ratio needed to cause the heavy lever to pull back up horizontal? Is there enough mechanical advantage with 2.53MA to lift using the gear ratio? I think there would be. 2.53 MA is a lot and there is only a small amount of gear ratio needed to make the heavy gear pull up horizontal.

This could be a working gravity wheel design!! I want to know if I am correct.

[rlortie]

Preoccupied,

I do not know if it would work or not, but I must admit that it is very interesting to watch. I have been glued to watching it for some time and wondering if it is possible to work around the faults and design the necessary mechanisms to test it.

The first thing that comes to mind is keeping the right heavy lever from hitting the axle and allow it to raise past 6:00.. Letting it hit the axle is a waste of good Ke that could be utilized. The second observation leads to a catch and release system to transfer the left weight from one lever to the the other and back to the ramp. Third flaw is connectivity between the two levers.

Having experienced extensive builds with right angle levers and knowing of one that would run in a living room when the washing machine went into its spin cycle is baiting!

You have my fullest attention on this, I do believe I will be doing farther research as soon as I get a backlog of submitters whittled down.

As for gear ratios; I do not see a reason to compound the issue, what you have can be accomplished with simple leverage.

Being a new member you may not be acquainted with my method of research. I throw the math and simulations out the window and use strictly hands on research. If it runs, then I let the math boys attempt to figure out why...

Ralph

[murilo]

Not clear is how that 'L' piece with the bigger pink ball is going to reach its right position.
I mean how the ball will reach the 3h position before lever rise back.
Something is hidden?
Good design!
Best!
M.

[preoccupied]

Thanks for your curiosity murilo. The L shaped bar's pivot is connected to the top pivot so that when the top pivot moves it applies force to the bottom pivot causing it to move too. I wanted the top pivot to be a ratchet so it will only apply force in one direction. The reason the top pivot would be able to move the bottom pivot is because the length of the lever on the top pivot is larger than the lever on the L shaped bar connected to the big weight. The top lever is larger because the right angle bar pointing down adds extra space downward to drop the weight after it falls. The ramp could be any size and the lever could be any size but it would need to drop onto something that can pick it back up and to do that a right angle bar is needed to maximize that distance.

[jim_mich]

This is interesting. You would need a 7:10 gear ratio for it to function properly. This give you about 0.4º of slope for the weight to roll down to the left. Initially the small weight has enough weight and leverage to start lifting the large weight. The small weight starts at 4.47 inches out with a 7:10 ratio = 1_wt × 4.47 × 7/10 = 3.129 torque force on left side with zero initial torque force from gravity on the right side. (See left picture.)

But soon the tables turn. At the end of the fall the left side has 1_wt × 2 × 7/10 = 1.4 torque force while the right side has 2_wt × 2.83 = 5.66 torque force. (See right picture.)

We might want to look at the possibility the moving weights could gain enough initial momentum to raise the large weight up to straight out. I have strong doubts that it could, but this reminds me of the "uncle's toy".

Here is the math:

Assume angle 'aº' is zero when Wt.2 is out to the right at 3 o'clock.
Thus when Wt.2 is straight down then aº = 90º.


Distance X0 = Cos(aº+26.56)×4.47 (Left weight falling)
Torque T0 = Wt1×X0×0.7

Distance X1 = Sin(aº+45)×2.83 (Left weight rising)
Torque T1 = Wt1×X1

Distance X2 = Cos(aº)×2.83 (Right weight falling or rising)
Torque T2 = Wt2×X2

If small Wt.1 = 1N and large Wt.2 = 2N then the falling weight torque from gravity will match the rising large Wt.2 when angle aº is equal to about 56.7º.

If we increase Wt.1 to about 1.35 then the torque of the two weights will balance at aº = 45º.

This is an unusual situation and could use a closer look.

Note that the torque generated by the weights does not depend upon the shape of the lever. A lever that has a right angled bend works exactly the same as a straight lever, assuming both weigh the same. Thus all we really need to know is the location of the weight mass.