Idea I had since elementary school

[preoccupied]

I'm planning to test levers torque using a torque wrench even though science says it won't have different torque. I have an experience that the right angle bar produces different torque. When I was jacking up a car the jack had a right angle bar for pushing and pulling. When I pulled on the inside of the right angle bar it was easy and when I pulled on the outside of the right angle bar it was difficult. My experience leads me to believe that the right angle bar produces mechanical advantage. I think it's possible that I'm right and the science encounters the right angle bar producing mechanical advantage but it doesn't interfere with other calculations so it is not noticed. If the right angle bar was purposely used on the jack to produce leverage then engineers might know about this. I'm probably wrong but as soon as this person I know finds his torque wrench I'm going to try to build a right angle bar and test it to a straight bar to the same position.

Keep me updated on further calculation with radiusshift11. I can't do that math because I can't remember the math I learned in High School. The windows calculator doesn't have sin cos or tan. I lost my old calculator that I used in High School math. I have covered the subjects of sin cos tan in school but I never took physics.

[jim_mich]

The windows calculator doesn't have sin cos or tan.

Sure it does. Just click 'View' and then click 'Scientific'.

[preoccupied]

On radiusshift11 what is about the average torque for all positions going down on the long lever before the heavy weight 'turns the tables' and balances it out. If the average is enough to push up the heavy weight maybe multiple devices connected to each other might be able to complete the task. The lever when it is all the way up has the most torque so when it is half way down a different machine's weight can start at the top and contribute to both machines when its weight falls on the long lever. If the average for both machines is enough to move their weights to the correct positions then one machine can reload and the other machine can wait until the first machine can reach the long lever again. I think the ability for the weight to move the heavy weight partially with on the long lever comes from the weight rolling down the ramp but the full distance of the ramp is not the full distance of the mechanism so more than one mechanism is needed to complete the task.

This could be a real perpetual motion machine!

EDIT

Also interesting to note is that if two machines don't manage to reload one machine a third machine's weight could be used because any amount of machines can wait to be reloaded and still be able to complete the task.

[AB Hammer]

preoccupied


It is an interesting idea but I can't see it lifting the moving weight back to the top. Once the weight has connected with the bent arm it becomes a part of the larger weight and may move a little but they will hand together. You will need something else to finish the movement.


Alan

[preoccupied]

AB_Hammer, the L shaped lever will lift the weight to the ramp according to Jim_Mich's math in his writing. I think Jim_Mich is right. Note that the lifting of the weight is not connected to both machines only the falling of the weight on the long lever is connected because the axle is a ratchet. Thank you Jim_Mich for your torque calculations!

[path_finder]

Dear preoccupied,
IMHO the low value of the inclined ramp don't allow the capture of any big amount of energy.
It could be at the same level than the 'drinking bird', said a tilt of a very light pendulum.
See here:http://en.wikipedia.org/wiki/Drinking_bird#How_it_works
and here: http://www.youtube.com/watch?v=_0mcGDAs3f8
The summary of the PE (potential energy) seems to be very low.
In any case many thanks for sharing your ideas.

[AB Hammer]

preoccupied

from the point of the weight added to the elbowed arm. Now draw a straight line from it to the other weight. Now look at where the fulcrum is. It will not stand straight up it will keep a lean. It will be a very easy test to do with coat hanger wire or wooden dow rods with some tape. Do this and see what happens and you will see what I am talking about.


But good luck

Alan

[preoccupied]

AB Hammer, I think you are making an assumption without proof. If you calculated the torque you would know if you are right or wrong about the L shaped lever not lifting to the ramp. I actually don't know how to calculate the torque. Jim_Mich knew how to calculate torque and that is why he gave an explanation of the long lever lifting the heavy weight partially. To me it seems like a heavier weight would be able to hold a lighter weight up on an angle if it is heavy enough. How heavy must the heavy weight be to lift the lighter weight on the L shaped lever to the ramp? Can anyone help me with this, please? I think AB Hammer is making an assumption.

If the L shaped lever can lift to the ramp it can do so on its own every time. It's the long lever that can only go partially. The idea is that if there is a second machine using the long lever on it that that part of the long lever where it has the most leverage on the top would be able to lift the heavy weight back up on the first machine so that the small weight can connect to the L shaped lever again on the first machine.

[preoccupied]

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

I should be able to figure out how much weight is necessary to lift Wt1 using Jim_Mich's formulas but I can't figure out how to properly use Sin and Cos using windows calculator. I found the scientific setting but whenever I use Sin it gives me 0. AB_Hammer has me worried that it is impossible to lift Wt1 to the ramp no matter how heavy Wt2 is.

[James.Lindgaard]

To illustrate I made this quick drawing in Paint. As you all should know a right angle has mechanical advantage and a square does not. When the first square moves it moves both the second square and the swastika. The second square might not be necessary because it is the red rods that are important. Note that the red rods are connected to the hands of the swastika which are blue and the corners of the squares. The main hypothesis is that the axle of the swastika can connect to another square and repeat the mechanical advantage again.

The rods are red. The swastika's axle is white. The square's axle is black. The hands of the swastika are blue.

preoccupied,
One of the reasons I have not commented on your idea is I have already built it. If you look at your middle diagram, going counter clockwsie, the arm at 9 o'clock can lean out creating an obvious over balance.
It did not develop sufficent force to maintain rotation.
Also, the swastika as it is known, before the 3ed Reich was a symbol that represented purity and power. it is quite anceint. This is why the Nazi's chose it. For it's meaning throughout history. They onle desecrated it.

[James.Lindgaard]

preoccupied,
I thought I would give you a better answer.
When the arm level with the axel swings out to an over balanced position, it's force on it's pivot point diminshes. This allows for counter rotation of the wheel.
When you consider the mass of the wheel minus one lever, it is mv-1l=f.
f needs to equal the mass of the wheel.
The weight moving to the over balanced postion needs to account for this. It is a force requirement in excess of the 3 weights as the frame work for the wheel is included in the counter rotation calculations.
Simply put, a poisitve or balanced force always needs to be maintained,
Even a small surplus of positive net force is not sufficient. The entire mass of the wheel needs to be considered when wondering, if I have mv-f, how much mass do I have ? And this is after accounting for the energy lost due to friction and parts moving.

James (aka P-Mo)

[preoccupied]

I'm happy you decided to look at my post James. I wish I could understand what you are talking about but as usual none of it makes sense. Do you have a college degree in anti-communication?

If you already built it does it work!?! The long lever has the most torque when it first starts so having multiple application of that first larger torque could cause one weight to be reloaded. If you can imagine it the two moving weights from two machines might not be enough so a third machine could be added. From the first two machines the first weight would be slightly closer to being reloaded than with just the first machine and the third machine's moving weight could be enough to cause the first weight to be reloaded. If one weight can be reloaded then the machine can operate under the force of gravity. Only one weight needs to be reloaded at a time.

[preoccupied]

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.)

The calculations say that it starts with 3.129 torque but then has a negative torque of 5.66-1.4=4.26 torque at the end (if it were to hypothetically reach there). If we add the small weight's torque with the second machines starting torque 1.4 + 3.129 = 4.258 we get the exact same torque. Interesting. I think that means two machines will take the reload a little over half way between the weight's current position and the destination and a third machine would take it all the way to the destination. I hypothesize that three machines connected together will cause a the mechanism to turn using the force of gravity.

[AB Hammer]

preoccupied

your approach a good idea but here is what I see. Bellow is a photo of a wheel with 2 weights on it 1- 4oz and the other 1-1lb 8oz. As you can see it does not make a full lift. No mater how much weight you add to the bottom you only get a little at a time. I am sure you are expecting your counter weight to make the balance and the drops off onto the ramp again.
Now then you have the problem of moving it. Your upper rocker counter weight becomes a problem in other ways. If your upper weights are to small you will not rock the lower weight. The pick up point is down at the end of the ramp so your counter balance is not enough to balance the now lower weights to lift it. It is a typical each change to help in one way you hurt the other.
Just think how much weight you will need just to rock the lower weight.

It is still worth a build due to the learning curve. A little over confidence helps keep you going.

[AB Hammer]

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.)

The calculations say that it starts with 3.129 torque but then has a negative torque of 5.66-1.4=4.26 torque at the end (if it were to hypothetically reach there). If we add the small weight's torque with the second machines starting torque 1.4 + 3.129 = 4.258 we get the exact same torque. Interesting. I think that means two machines will take the reload a little over half way between the weight's current position and the destination and a third machine would take it all the way to the destination. I hypothesize that three machines connected together will cause a the mechanism to turn using the force of gravity.

You may be calculating it a bit wrong. If you get half more the next one will only give you half that more, thus for example the large Costa wheel in France.