Musing

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Kirk
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Musing

Post by Kirk »

Amass of 2 falls 16 feet and strikes a 2:1 lever with a mass of 1 on the long end. The mass of 1 departs at twice the velocity of the mass of 2 rising 4 times as far. Although it weighs half as much 4 times the height gives you twice the potential energy of the original condition.

Room in there for losses and maybe a bit more.


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re: Musing

Post by pequaide »

Yes; but the larger mass has to give all of its motion to the small object and stop (without friction or hitting something), and in the lever it will not. But the larger mass in the yo-yo despin device will stop and give all its motion to the smaller mass.
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re: Musing

Post by Tarsier79 »

Been there, done that. The large weight does not give all its momentum. It gives all its energy, shown by its complete transition to a stationary mass and producing a near COE.

Do the experiment, video and share, then be called a liar for not getting a gain, eh peq.
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re: Musing

Post by Fletcher »

Hi Kirk .. long time no hear.

That's why we were wondering about the use of tapered storksbills in another recent thread. Ordinary storksbills give an increasing acceleration/deceleration curve but then it starts to diminish and peter out.

We actually want the load acceleration to exponentially increase whilst stopping (or near stopping) the falling driver mass so that effectively all momentum from the driver goes into the load.

Then you would see what you predict. IOW's a violation of CoE.

I'd suggest you'd need to find the right kind of mechanical force amplifier to do that (or lever as you prefer). Pequaide is convinced that the yo-yo de-spin device and trebuchet have the ability to do that.
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re: Musing

Post by Kirk »

In my example the lever is infinitely rigid and mass-less. Matching is essential and the mass of the lever enters into this in the real world.

I suggest the experiment be retried with this in mind. Since the theoretical return is 2 practical results should be 1.4 or above. Lever has to be stiff or such that the reflected wave arrives at the driven mass at the right time.

Start horizontally operating to remove g measuring whilst increasing velocity and it should be good at low levels. The rigidity of the lever is a problem. The forces are high.
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Re: re: Musing

Post by Kirk »

Fletcher wrote:Hi Kirk .. long time no hear.

.
Health was a train wreck. I shouldn't be here for several reasons.

Good to hear from you again. Thought you guys would have solved it by now.

The math is wonkie. kinetic energy is misleading. I use momentum for my pencilings as it is real.

We are stuck in our framework.

If you can design the mechanical equivalent of a transformer and match impedances that would be interesting. Power in some formulas is proportional to V squared as mechanical energy is proportional to V squared also but velocity instead of voltage.
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re: Musing

Post by jdelaney »

Kirk,
I know how health problems go, glad to hear you're better.
I got to thinking about what you posted and checked on something quick like. It was basically how scissors move. And that's with one lever rotating. The opposite lever does not have a fixed point on the base. This allows the other side of the scissors to move towards the fixed side. A small detail but quite important.
With Mt 26, it shows 1/2 of a scissor with a wheel weight. And with Mt 20, it shows a long lever basically having a pulley at it's fulcrum. The drawing shows those 2 things put together.
As the wheel rotates towards 90 degrees ATDC (3 o'clock), the dimension on the drawing uses a 10 inch long lever rotating about 10 degrees (a 2 inch drop). This moves the weight on the scissors outwards from about 2 3/4 inches to over 11 inches. That's a lot of over balance.
And to retract the weight ? The long lever that would be moving above 90 degrees BTDC (about 9 o'clock) would drop it's 2 inches. This would require the weights to work together as well.
I guess this, if it worked then might show, well, I took your weight dropping 16 feet and converted it to leverage. I hope you don't mind. Of course, if something like what I drew can actually work, then you'd be a part of it's realization. Would be pretty good timing on your part.

Jim

p.s., this is drawn with 3 inch levers on the scissors so it's about a 3:1 ratio.

edited to change ratio to about 3:1. this would actually be the pulley on the long lever would need to have 3 times the diameter as the one on the lever for the scissors. With actual builds, different ratios could be tried.

edited to add; modified the drawing to show about what it would look like with 4 sets of scissors. The hub (mounts) would form a square for the most part.
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Kirk
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re: Musing

Post by Kirk »

Too complicated for this late at night. The problem as I see it with schemes like a scissor is the velocity of the object is dropping and when you reach the halfway point its velocity is too low for the continued acceleration. The rigid lever is a modified elastic collision.
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re: Musing

Post by Fletcher »

Here's the sim of your idea Kirk. The Yellow 2kg Driver falls from 2 meters height (GPE = 40 J's). The Green 1kg Load rises just short of 4 meters gaining about 33.75 J's of GPE.

This is with the lever set to 1 gram mass; and all parts have elasticity set to 0.95. N.B. at elasticity at 1.00 it is a perfect spring analogue and because parts rest and collide they behave weirdly because 1.00 is not achievable in real world.
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re: Musing

Post by pequaide »

Kirk used Newtonian math and it is four times as high; or 8 meters. Your sim is programed non-Newtonian.

You will have to trust your own abilities Kirk.
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re: Musing

Post by Fletcher »

You forgot about inertia of the 1kg at 2 meters pequaide.
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Re: re: Musing

Post by jdelaney »

Kirk wrote:Too complicated for this late at night. The problem as I see it with schemes like a scissor is the velocity of the object is dropping and when you reach the halfway point its velocity is too low for the continued acceleration. The rigid lever is a modified elastic collision.
Kirk,
With moving a weight outward, acceleration is easy because a weight is moving perpendicular to gravity, little resistance.
As with your original post, it was about accelerating a mass upwards using leverage and force. With scissors we'll say lifting. And since a mass develops force as it drops, we can say force into velocity.
And with a 2:1 ratio, lift should happen. The difficult part might be that with a scissor, if you consider the pivot in the middle as being lifted, then everything else is an upward cascading effect.
What's shown is an illustration. Larger pulleys would be needed.

Jim Lindgaard
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re: Musing

Post by jdelaney »

@All,
The drawing is a basic test that can be tried if anyone is willing. I have drawn it to scale, 1cm equals 1 in.
The long lever might only drop 4 in. before it is stopped because the scissors closed. The weights on both the long lever and being lifted by the scissors are the same mass.
The pulley / roller at the top center of the scissors is used only to guide the line doing the lifting, it does not change any ratios.
And if the weight on the long lever is able to be repositioned, then by moving it, the ratio of it's drop to the lifting of another weight will change.
I would try this but do have 2 builds that I'm working on and do need to finish those before building anything else.

Jim

edited to add; with the end view, it shows the lever for the scissor being supported by an axle. and of course, the axle has a nylon bushing or a bearing it rolls in inside of a track. Think of a sliding door that is supported from the top like with a closet, etc.
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Re: re: Musing

Post by AB Hammer »

Kirk wrote:Too complicated for this late at night. The problem as I see it with schemes like a scissor is the velocity of the object is dropping and when you reach the halfway point its velocity is too low for the continued acceleration. The rigid lever is a modified elastic collision.
Greetings Kirk

Nice to see you here again.

There is more than meets the eye about scissor jacks. They work better side to side than up a down. The scissor jack is known as a compound lever. To lift a pound by squeezing one section and the other weight is 4 more sections away to the moving weight to lift it will have to be 4 times plus the weight to compensate for the materiel of the scissor jack just to equal. 5 times to lift it. The traditional ways we tend to work with scissor jacks do not work in our wheels. But there are other possible ways to work with them. The longer lever as shown before my post helps to lift but is it worth the trade off in the wheel even though the change in his last post is an improvement. There are also some other pictures of other ways to use the jacks in the albums here. To many trade offs tend to leave us standing still.

Here is one that I found in a quick look.

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http://www.besslerwheel.com/forum/download.php?u=2409
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re: Musing

Post by jdelaney »

I made a major mistake in the drawing. To simplify, a 26 in. long lever dropping 10 degrees drops about 4.5 inches. It can rotate a lever 8 inches long 1.5 inches upward. And with scissors, as AB Hammer correctly pointed out, scissors allow for a multiplying of the distance with a single movement.
And if using more levers in a scissor and making them shorter, then the amount of force that can be generated in this increases exponentially.
It's just that by trying something perpetually, the lever that expands scissors can not be the one that retracts them. What this would require is that weights work together to expand and then to retract scissors.
And with what I just mentioned in this post, it would probably be a good starting point to understanding how this could be applied to Bessler's drawings and some of the descriptions he gave. Myself, I am convinced that Bessler knew more than one way to skin a cat.

@AB Hammer, I was giving you the opportunity to show that you are capable of working with other people. A simple force calculation at a 3:1 ratio would say that 20 in. lb. / 3 = 6.7 in lbs. of force. Is that enough to lift a 1 lb. weight ? According to what you just posted, it is not.

edited to add; as for a demonstration using a 26 in. and a lever 8 in., with scissors, the lever doing the lifting would need to be on the outside of the scissors on both sides so it would be lifting the point where the scissors have a common pivot point in the middle.
all this means is that the 26 and 8 inch levers would be pivoting on the same fixed point that the scissors do. and this is about as simple as it gets.
of course, for testing purposes, the 8 inch lever can be shortened to 4 inches. This means that the levers in the scissors would be 8 inches from where the center of rotation for the levers are.
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