preoccupied,
Do you mean a bellows like this ?
I am also building a 4 bellow/weight combination. I'm probably doing it out of boredom. Not much to be learned from such basic engineering/wood working. Of course, having worked with metal, I do find wood to be much more difficult. You can't beat it with a hammer.
It is some pretty basic math. The bellow that is open holds twice as much water as the one that is closed.
I should have the outer rims finished this weekend but am working on the next build. This means it will take a month or more.
Bellows
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justsomeone
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re: Bellows
Yet another sock puppet? Hi Jim.
. I can assure the reader that there is something special behind the stork's bills.
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preoccupied
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re: Bellows
The red lever is 23.81 lenth because it's going to fall 22.5 degrees from its horizontal position and have 22 width and 9.11 height. It will compress the bellows that is a 45 degree bellows at a 1:1 ratio. Two weights fall, one at a time, applying a minimum of 22 force to the maximum 18 force required to lift the counter weight.
The counter weight is 36 weight and is reduced by the weight of the column of water to 18 at its maximum resistance. So the 22 force applied by the lever will cause this to be an over unity lever.
Just out of curiousity if this were only gear ratios, there would be a column of weights and I don't see the need for two columns of weights. The 9 units of weights holding two weights each would weigh 18. The counter weight octagon pink weight would have to weight 18.1 to lift the weight up one unit. If an 1/8th of the circumference of the gear used to lift the weights from the pink octagon is one unit and the same size gear is on the lever, then the two weights would fall 22.5 degrees each at the length of the lever minus the radius of the gear 22-1.3=20.7, 20.7 force must lift 18 weight. So this arrangement seems to work with weights and gear ratios also.
I'm concluding that it's not the bellows that is making this calculation work but my calcualtion on the use of two weights falling horizontal position to 22.5 degrees to lift a row of weights stacked two together, that has a counter weight that lifts them up. The counter weight is lifted by the lever falling. At this point I'm not sure that the counter weight is necessary. I mean it weights 0.1 more than the row of weight to lift up.
The counter weight is 36 weight and is reduced by the weight of the column of water to 18 at its maximum resistance. So the 22 force applied by the lever will cause this to be an over unity lever.
Just out of curiousity if this were only gear ratios, there would be a column of weights and I don't see the need for two columns of weights. The 9 units of weights holding two weights each would weigh 18. The counter weight octagon pink weight would have to weight 18.1 to lift the weight up one unit. If an 1/8th of the circumference of the gear used to lift the weights from the pink octagon is one unit and the same size gear is on the lever, then the two weights would fall 22.5 degrees each at the length of the lever minus the radius of the gear 22-1.3=20.7, 20.7 force must lift 18 weight. So this arrangement seems to work with weights and gear ratios also.
I'm concluding that it's not the bellows that is making this calculation work but my calcualtion on the use of two weights falling horizontal position to 22.5 degrees to lift a row of weights stacked two together, that has a counter weight that lifts them up. The counter weight is lifted by the lever falling. At this point I'm not sure that the counter weight is necessary. I mean it weights 0.1 more than the row of weight to lift up.
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preoccupied
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re: Bellows
The circumference of a circle to find the gear that can lift a weight vertically one unit when falling 45 degrees is, C=2(3.1415)R. To find the radius I need to divide 6.28 by the circumference that I want. I want an 8 circumference so that 45 degrees turn moves one unit. 8/6.28 about equals 1.3.
The goal is to lift the 18 weights one unit up using the long lever. The weights sit together in groups of two nine units up and down. The lever falls from a horizontal position to 22.5 degrees. The lever length is about 23.5 because of the 22.5 degree angle and the 21.72 width because of the 9 height. However the shortest the lever will be is 21.72 width from its fulcrum so that's what's important.
The gear used to lift the weights is 1.3 radius so that one unit is moved per 45 degrees turn. The lever is therefore about 21.72-1.3=20.42 length of force and the weights are 18 weights. 20.42 lifts 18, it's over unity!!!!
In my ms painting I did draw pentagon counter weight but it is not necessary.
When I first was looking at this arrangement I was looking for a way to make my bellows more efficient but now I realize when the two weights work together at an angle they can have a long enough lever to be over unity. It's the over unity lever. I've found it guys. LOOK! It's right there. Johann Bessler may have used this for his gravity driven wheel.
Now that I've found this, you know what I want. Bring me all of the prettiest female lovers that I can possibly want then more and a big pile of cash. Thank you very much.
edit
I noticed that my ms painting is not drawn to scale which makes the grid kind of silly. I'll work on that. My presentation is poor then I guess but this is still an extraordinary find.
The goal is to lift the 18 weights one unit up using the long lever. The weights sit together in groups of two nine units up and down. The lever falls from a horizontal position to 22.5 degrees. The lever length is about 23.5 because of the 22.5 degree angle and the 21.72 width because of the 9 height. However the shortest the lever will be is 21.72 width from its fulcrum so that's what's important.
The gear used to lift the weights is 1.3 radius so that one unit is moved per 45 degrees turn. The lever is therefore about 21.72-1.3=20.42 length of force and the weights are 18 weights. 20.42 lifts 18, it's over unity!!!!
In my ms painting I did draw pentagon counter weight but it is not necessary.
When I first was looking at this arrangement I was looking for a way to make my bellows more efficient but now I realize when the two weights work together at an angle they can have a long enough lever to be over unity. It's the over unity lever. I've found it guys. LOOK! It's right there. Johann Bessler may have used this for his gravity driven wheel.
Now that I've found this, you know what I want. Bring me all of the prettiest female lovers that I can possibly want then more and a big pile of cash. Thank you very much.
edit
I noticed that my ms painting is not drawn to scale which makes the grid kind of silly. I'll work on that. My presentation is poor then I guess but this is still an extraordinary find.
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re: Bellows
One positive is that your drawings always have a built-in bindle stick.preoccupied wrote:My presentation is poor
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preoccupied
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re: Bellows
ED, My bindle stick is supposed to be a perpetual motion machine man....
There are 18 weights being lifted vertically in nine spots where two weights are in each spot. They are lifted by a lever attached to a gear that pulls up the weights.
The gear is 8 units around so that 45 degree turn would lift the weight. To find the radius that would be subtracted from the lever, I take C=2(3.1415)R and put C=8... 8/6.28= about 1.3.
The lever falls 9 units to a 22.5 degree angle. The length of the lever is therefore 23.5 as I found on a trig calc. It is also 21.72 at its lowest position. It's lowest position will determine if it is an over unity lever.
My mistake in my previous post was cancelling out lever. That can't be done.
18 weight * 1.3 lever = 23.4. I simply kept it at 18 and cancelled out the lever difference. OOPS! The lowest position on the long lever is 21.72 so this is not an over unity lever. I'm not giving up on this.
There are 18 weights being lifted vertically in nine spots where two weights are in each spot. They are lifted by a lever attached to a gear that pulls up the weights.
The gear is 8 units around so that 45 degree turn would lift the weight. To find the radius that would be subtracted from the lever, I take C=2(3.1415)R and put C=8... 8/6.28= about 1.3.
The lever falls 9 units to a 22.5 degree angle. The length of the lever is therefore 23.5 as I found on a trig calc. It is also 21.72 at its lowest position. It's lowest position will determine if it is an over unity lever.
My mistake in my previous post was cancelling out lever. That can't be done.
18 weight * 1.3 lever = 23.4. I simply kept it at 18 and cancelled out the lever difference. OOPS! The lowest position on the long lever is 21.72 so this is not an over unity lever. I'm not giving up on this.
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Re: re: Bellows
Justsomeone,justsomeone wrote:Yet another sock puppet? Hi Jim.
Sadly, Scott's green dot thing has allowed for cronyism to take over his forum. As for me, I think you are a joke.
re: Bellows
@All,
Here's a pic of what I'm building. it's Mt 127. What I am not showing is what I am really working on.
I have bought some wood to make new rims. Butchered the ones I have trying to true them up. Still, not sure it'd be worth my while to prove a Bessler drawing actually works as drawn.
Here's a pic of what I'm building. it's Mt 127. What I am not showing is what I am really working on.
I have bought some wood to make new rims. Butchered the ones I have trying to true them up. Still, not sure it'd be worth my while to prove a Bessler drawing actually works as drawn.
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preoccupied
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re: Bellows
brian3051, That is a unique way to lift water up vertically. The weight on the lever is connected by a rod to an L shaped lever that opens and closes the bellows. The calculations necessary to know if it would work is beyond my scope but I recommend you measure twice and cut once. If you can put it all on paper you might not need to build. Or if you build you would have astounding support for it. Can you show me your math?
I think the pressure on the bellows would be pretty constant. When the bellows is opening it will have more pressure than when it is closing but they would both be pretty similar. I see in the mt 127 that the lever starts where the L shaped lever that moves the bellows is and ends at the weight on the end of the long lever. I think this is perfectly positioned in the drawing by Bessler because he is not looking to have a long stroke to move the L shaped lever but still wants some distance for the weight to be placed on. If you see what I'm talking about, then you know that I mean that the long length of the lever attached to the weight is not supposed to produce extra leverage but a shorter stroke and the actual length of the lever starts at the L shaped lever at the bellows and ends at the end of the weight. Or at least that is how I see it in mt 127. I will elaborate more on this if you didn't get what I mean. Just ask.
Please show me your math. I don't think it's as simple as saying there is twice as much water on one side of the bellows than the other like you said. That is not good enough to satisfy my curiosity.
Are you a James? And have you been on the forum more than once?
I think the pressure on the bellows would be pretty constant. When the bellows is opening it will have more pressure than when it is closing but they would both be pretty similar. I see in the mt 127 that the lever starts where the L shaped lever that moves the bellows is and ends at the weight on the end of the long lever. I think this is perfectly positioned in the drawing by Bessler because he is not looking to have a long stroke to move the L shaped lever but still wants some distance for the weight to be placed on. If you see what I'm talking about, then you know that I mean that the long length of the lever attached to the weight is not supposed to produce extra leverage but a shorter stroke and the actual length of the lever starts at the L shaped lever at the bellows and ends at the end of the weight. Or at least that is how I see it in mt 127. I will elaborate more on this if you didn't get what I mean. Just ask.
Please show me your math. I don't think it's as simple as saying there is twice as much water on one side of the bellows than the other like you said. That is not good enough to satisfy my curiosity.
Are you a James? And have you been on the forum more than once?
re: Bellows
preoccupied, I am James/Jim.
With what I have seen of your math, it reminds me of when I first started working on different idea's.
With something like this, it will be twice as much water on one side as the other.
If there is 3 lbs. of water in both bellows (1.5kg's if you don't live in America ),
if 2 lbs. of water are in the bottom bellow (open) and 1 lb. in the top (closed), to open the top bellows will require 3 lbs. of force times distance. And if the distance is 2 inches, then 6 inch lbs. of force is all it would take.
And if both levers working together have the force of one lever's total torque, then levers 24 inches long with a 1 lb. weight will generate 24 inch lbs. of force, 4 times the required force to open the top bellow.
Bessler did say the weight shot out and with a set up like this, it will shoot out of the bottom bellow and up into the top bellow.
This kind of goes to lifting 4 times the weight 4 times the distance. And to show that Mt 127 works perpetually as drawn would be as close as possible to proving that Bessler did in fact build work wheels as he has written about.
It might take you getting a little used to this one thought, but how much force does it require to suspend a weight in air ? If it takes only 1 lb. of force to suspend 1 lb., then work is measured by how much the bellow moves when it opens and not the distance water travels.
With what I have seen of your math, it reminds me of when I first started working on different idea's.
With something like this, it will be twice as much water on one side as the other.
If there is 3 lbs. of water in both bellows (1.5kg's if you don't live in America ),
if 2 lbs. of water are in the bottom bellow (open) and 1 lb. in the top (closed), to open the top bellows will require 3 lbs. of force times distance. And if the distance is 2 inches, then 6 inch lbs. of force is all it would take.
And if both levers working together have the force of one lever's total torque, then levers 24 inches long with a 1 lb. weight will generate 24 inch lbs. of force, 4 times the required force to open the top bellow.
Bessler did say the weight shot out and with a set up like this, it will shoot out of the bottom bellow and up into the top bellow.
This kind of goes to lifting 4 times the weight 4 times the distance. And to show that Mt 127 works perpetually as drawn would be as close as possible to proving that Bessler did in fact build work wheels as he has written about.
It might take you getting a little used to this one thought, but how much force does it require to suspend a weight in air ? If it takes only 1 lb. of force to suspend 1 lb., then work is measured by how much the bellow moves when it opens and not the distance water travels.
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preoccupied
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re: Bellows
As you can see in ms painting the water travels up from bottom to top a distance of 14 units. At the bottom the pressure on the bellows is the distance of the lever times 14, 7*14=98. Just above that is 13*6=78, and then 12*5=60 and so on, 11*4=44, 10*3=30, 9*2=18, 8*1=8, and then the new bellows on top's resistance starts, and it will be 1*7=7, 2*6=12, 3*5=15, 4*4=16, 5*3=15, 6*2=12, 1*7=7. And 98+78+60+44+30+18+8+7+12+15+16+12+7= 405.
It takes about 405 units to move the bellows at its greatest resistance possible. The next step in the calculation is determining how the levers apply pressure, so that we know how much leverage the weight will have and then we can determine how much the weight will weigh. When we know how much the weight will weigh then we can see how the weights position will effect the balance of the wheel. If it's a super heavy weight and is acting as a counter weight it might prevent the rotation of the wheel.
It takes about 405 units to move the bellows at its greatest resistance possible. The next step in the calculation is determining how the levers apply pressure, so that we know how much leverage the weight will have and then we can determine how much the weight will weigh. When we know how much the weight will weigh then we can see how the weights position will effect the balance of the wheel. If it's a super heavy weight and is acting as a counter weight it might prevent the rotation of the wheel.
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re: Bellows
preoccupied,
One thing that helped me was in simplifying the math. Circumference is Pi2R. If the 12 o'clock CoG (radius) and the 3 o'clock CoG (radius) equal x + y / 2 * Pi = z, then z is an approximation of the distance a weight travels on the over balanced side.
If the same math is used on the under balanced side, then net force becomes known as a percentage of force.
Jim
edited to add; preoccupied, with Mt 21, if you were to calculate the average path of 2 opposing levers, their weights would travel the same distance.
Yet if each lever were 18 inches long and had a 1 lb, weight at it's end, that would be 36 inch lbs. of force not being put to use. Some food for thought.
One thing that helped me was in simplifying the math. Circumference is Pi2R. If the 12 o'clock CoG (radius) and the 3 o'clock CoG (radius) equal x + y / 2 * Pi = z, then z is an approximation of the distance a weight travels on the over balanced side.
If the same math is used on the under balanced side, then net force becomes known as a percentage of force.
Jim
edited to add; preoccupied, with Mt 21, if you were to calculate the average path of 2 opposing levers, their weights would travel the same distance.
Yet if each lever were 18 inches long and had a 1 lb, weight at it's end, that would be 36 inch lbs. of force not being put to use. Some food for thought.
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preoccupied
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re: Bellows
the weight has 6 units of leverage. 405/6=67.5. The weight weighs 67.5. It is displaced two units to the left when turning clockwise. I will say it's about 1 unit out of balance because of the weight because the weight also falls flat on the flip side. It's an estimation. I just want to move on to the water in the bellows.
There is no water in the bellows on the left side only on the right side. It has 0.28*1 and 0.56*2 and 0.84*3 and 1.12*4 and 1.3*5 and 1.58*6 and 1.86*7, so 0.28+1.12+2.52+4.48+6.5+9.48+13.02=37.4. Okay so right now the way I've drawn out mt 121 the weight is off balancing the wheel because the water weighs 37.4 force and the weight counter weighs 67.5. I multiplied the distance on the lever times the water weight. It's all being measured in water weight per square unit.
Jim don't simplify. Do it the way I'm doing it. Or the way someone else does it that better explains or knows more about the bellows.
Jim, how do I set this up so that the weight isn't counter balancing so much force? Do I have the lever set up wrong?
There is no water in the bellows on the left side only on the right side. It has 0.28*1 and 0.56*2 and 0.84*3 and 1.12*4 and 1.3*5 and 1.58*6 and 1.86*7, so 0.28+1.12+2.52+4.48+6.5+9.48+13.02=37.4. Okay so right now the way I've drawn out mt 121 the weight is off balancing the wheel because the water weighs 37.4 force and the weight counter weighs 67.5. I multiplied the distance on the lever times the water weight. It's all being measured in water weight per square unit.
Jim don't simplify. Do it the way I'm doing it. Or the way someone else does it that better explains or knows more about the bellows.
Jim, how do I set this up so that the weight isn't counter balancing so much force? Do I have the lever set up wrong?
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Jim Williams
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re: Bellows
How are devices using bellows anything more than overunity devices in reverse? They are both always dependent on one side always being heavier, (or lighter) than the other.
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http://worldwide.espacenet.com/publicat ... cale=en_EP
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Last edited by Jim Williams on Thu May 08, 2014 4:27 pm, edited 1 time in total.
Re: re: Bellows
An opposing lever. There will be a small amount of imbalance from one weight being out of balance. The shift in water if greater, should allow it to work.preoccupied wrote: Jim, how do I set this up so that the weight isn't counter balancing so much force? Do I have the lever set up wrong?
With how I build Mt 127, water will be in both bellows. The bellow that is resistance will have 1/2 the amount of water as the over balance.
Anyway, with the builds I am doing, using inch lbs. to understand the work I will need a lever to do will work just fine. This way, if this build works, then maybe others will build one.
Jim, I don't quite get this "How are devices using bellows anything more than overunity devices in reverse?".
It will be an over unity device, I'm not sure what you mean by in reverse. I think Bessler when writing about the work his wheels did was to give people an idea of how much power or velocity his designs could generate.