Idea I had since elementary school

[preoccupied]

AB Hammer,
If the squares are separated but are connected to the same one gear by connecting them all by their corners does it change the gear ratio or just add friction? One gear to turn them all and 4 gears to output some force. If the gear ratio doesn't change and it only adds friction what other forces are added that will reduce the movement? Because if only friction and gear ratio then there would be extra energy created. If this were more than 4 gears/levers being turned and weights were pulling the extra levers this would be a wheel that could turn in either direction just by pushing it into motion because one arm extends and on the other side one arm descend and it's reverse for the opposite direction.

In the picture the thick line is the axle and the four lines connected to the corners are connecting the 4 gears so that they move together. The gear that moves them would have an extended arm too but it is not in the animation. The big gear's arm which is not in the animation would touch just past the shortest part of the long levers so that it only pulls one lever at a time.

[preoccupied]

In the picture the squares are connected by four lines and an axle but the four lines would hit the levers when it turns so they can't be there on the corners of the squares. The four lines would have to be on the sides of the square instead of the corner. This would make the gear 12.56 instead of 17.76. The bigger gear is turning the smaller gear so the smaller 12.56 necessary to connect multiple squares would be a different more resistant gear ratio. But at that gear ratio of 12.56 any amount of squares can be added. If more squares are added what kind of resistance is created? I hope someone understands the science behind what I am asking for because I don't. Please tell me.

[AB Hammer]

preoccupied

Don't you think the rods from each corner of the square is going to hit the axle? You can not afford an axles but behind each square and will have to be an under axle connected by gears or drive chain, to allow them to spin with the rods. I think you need better diagrams to show your intent.

[Tarsier79]

AB, It is not the levers hitting the axle that is the main problem, because that can be rectified, it is the principle of the leverage, and where he thinks it is developing a gain in movement x power that is the concern.

[AB Hammer]

Tarsier79

That is the reason I asked (I think you need better diagrams to show your intent.) I can't see any advantage. I just can't get this do nothing device out of my head with this idea looking at it.

[Axelf]

@preoccupied
no ramp, this is wrong!

all your Gif animations are incorrect, please a little bit
logical thinking ....
a weight on one side, can only lift a weight on the other side, several levers can not increase or double.
the lever forces are always the same, but the distance you can
change, shorten or lengthen.
physics says force * distance, in special cases is not true,
that's you need to looking for.

[preoccupied]

I'm having some difficulty with my animation program inkscape. It makes everything an object and I guess the program can do a lot of tricks when it is operating properly but it isn't very helpful with drawing out things. For example it can't make circles or square shapes that are lines. Paint is worse because I can't move stuff around without the entire selected square overlapping over where it is placed.

I was looking at what I wrote and what I drew and I realize that I can't connect arms to new squares and levers because the levers pass all the way across the square so any arms attached to multiple squares would hit the levers when they are moving. I can't even attach an axle under the squares that would connect to all the squares because then the big gear would hit it. The big gear is surrounding the square using its arms and the levers cross over its own square when it turns so nothing can reach out or in and nothing can pass through the squares. No way in and no way out. Maybe there is no way to connect multiple squares and that is why this won't work. Because it seems like it would create extra force if multiple squares could connect to the one large gear.

The only way I can think of to connect the squares is to have a hollow axle and have one axle on the outside and one axle on the inside. The axle on the inside would have to be the multiple square's axle because it would have to reach outside the large gear that is surrounding it to give force to something else. The large gear would be on the outside of the hollow axle so it would have to be turned by the wheel of the gear instead of the axle. The squares would not be able to be the outside axle of the hollow axle because they are surrounded by the large gear's arms which prevent anything from getting out of the square except through the axle. Since the large gear is on the outside it can be turned on the outside of it using another gear attached to it.

[Axelf]

this is not a trick, you need only understand the law of the lever

[preoccupied]

What if there is a shovel with two bars and two heads but one handle? That would be just like there being two squares with four levers on it in the pictures. A force is applied to the hand of the shovel and it distributes evenly to both bars and to the heads of those bars which are the levers. Each bar would get half the input and both heads of the shovel would do their mechanical advantage. If the mechanical advantage is 1.16 like in the square and levers then lets say input =4, .5(4)*1.16=2.32 and .5(4)*1.16=2.32 and 2.32+2.32=4.64. If one bar and one head on the shovel were to use 1.16 MA then input 4 would be 4, 4*1.16=4.64. I guess the square with levers, no matter how many more squares and levers there are, won't create more MA than it already does because the input force is divided equally to all the levers. Alexf is right. It would have been super nice if someone had explained why I was wrong instead of me doing calculations myself.

[preoccupied]

How much distance can be gained by trading gear ratio for more ramp? There would be a .4 degree incline on the top ramp according to Jim_Mich so there could be a .4 incline on the bottom ramp too. That's .8 incline calculated against the distance of the ramp. There would be less gear on the long lever because it doesn't reach down as far; it only reaches to the bottom ramp instead of the right angle bar. The gear ratio should be over twice as resistant so instead of 7/10 it might be 3/10 gear ratio. I hypothesize that the length of the long lever can be more than five times its usual size and use around a 3/10 gear ratio and that would be able to create an overbalanced gravity system.

[AB Hammer]

preoccupied

Lets start by I like the basic idea of you rocker device, but there is several things wrong.

from the rock of the longer track arm you only twist 1/2 as much as the twist of the lower elbow unit. It would have to be geared in a way to do this which will also make the longer track arm to be twice as long to compensate then you would have another gearing problem so no gain. lets not forget that the elbowed arm will not do what you think it will do.

There are going to need some major changes to get this basic idea going. I will work up the drawings if you wish. The ways I look at these wheels and devices is it is a machine that don't work. So what has to be done to make it work. Once a basic idea is done that is where the real work starts. what changes and the best way to do a build. The only part of your device that I like after my complete evaluation is the long track arm. This is the only part I would keep to give you an idea. But I will also say at best it IMO will only make a good desk top toy even with my changes.

Alan

[preoccupied]

Any drawings and calculations would be appreciated, AB Hammer!

You're right that the right angle bar won't go all the way up but it will go most of the way up. There will be a slightly lower starting point for the top ramp but not much lower and that is good that it's not much lower. First calculation is how high can the right angle bar reach? Second calculation is hypothetical Distance A of a ramp and another hypothetical Distance B. Lets say Distance A is 5 times the length of the long levers normal length of about 4.47 inches or 22.35inches. Lets say Distance B is 10 times the normal long ramp or 44.5 inches. The incline would be about .4 degrees and that would double because there is a ramp on the bottom and a ramp on the top. The calculation is looking for the distance from the axle and how far down the ramp has gone to reach the distance of the longlever.

I would calculate the gear ratio but I can't. If someone can calculate the gear ratio for me that would be great. If we have the gear ratio we can calculate the torque and see where the right angle bar would be given a certain distance of the long lever. Somebody would be able to calculate the torque I mean because I can't. I can only calculate torque on a simple lever.

EDIT

Using this to calculate,
http://www.carbidedepot.com/formulas-trigright.asp

With a long lever of 22.35 inches the distance from the axle is 22.349 inches and the distance down from the axle is .156 inches. With a long lever of 44.5inches the distance from the axle is 44.4989 inches and the distance down from the axle is .31inches.

Now we need to know the gear ratio to see how significant the change in distance will be. It looks small but I can't know for sure. If the change is small then the long lever can be super long and that is VERY good for the hypothesis.

[Tarsier79]

You will find, the longer the lever, the less angle it rotates, the greater the gear ratio has to be...

[AB Hammer]

preoccupied

PM sent

[DrWhat]

Nice concept, good creative thinking.

You'll need a catch to keep the long lever in place whilst the ball rolls along the bottom track and then the catch is released when the ball attaches to the short arm.