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Author Topic: Gravity seems to yield unequal results.  (Read 11994 times)

Offline isodecryptor

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Gravity seems to yield unequal results.
« on: February 18, 2014, 11:10:09 AM »
I have been working on something for over 2 years. It is a strictly mathematical concept but I have had a good deal of college and this has bugged me for over 2 years, mathematically. My idea is similar to what the old grandfather clocks did. I will try to do my best to explain and also show any mathematical proof as needed.

The idea stems from a very simple concept that has to do with tension. A quick scenario includes two pole masses that are solid and connected to the earth. If I have a rope that connects the two poles with a mass, say 500kg and an angle of 5 degrees, I can easily find the tensions t1 and t2 in the rope. (Symmetrical)

This ends up being the same for any angle. When summing the forces in x direction with a symmetric setup, I get

T1cosx-T2Cosx=0  So I conclude that t1=t2, after crosssing off the cosine.
Now for the y I get sum of y forces is t1sinx+t2sinx-mg=0

Solving for either t1 or t2, both their own tensions, although equal, I replace either t1 with t2 or vice versa. So for t1, I get

Tension=T1sinx+T1sinx-mg. Adding both terms leads to 2T1sinx-mg=0 solving for T1=mg/(2sinx) (notice I am only solving for t1, even though both values are equal.

I get a function mg/(2sinx). Now lets say the mass is 500kg. So I have T1= 500*(9.8)/2sinx , so for any angle in this symmetrical setup, I could find both T1 and T2(Both are equal but their own values) by plugging in some angle for x in T1 or T2 =500*9.8/2sinx. In the case that the angle is smaller, this number starts to approach infinity, because this function is actually cscx, once you remove the constants, 500*9.8/2 cscx. Graphing cscx has interesting asymptotic behavior, which is the basis of my theory. Now, I thought good and hard about trying to find the work that could be done with such a system. I have considered many factors, and mathematically have yet to see any flaws with my theory. Now, if I have a smaller angle, my tensions grows very quick, and as I approach pi/2, my tension becomes exactly half the weight of the mass that is hanging. Now, I have two tensions which are pulling on two poles. If these poles had moving parts( such as pully's) I can in-vision calculating the work done on these two pivoting points. Please graph cscx to see that my reasoning is exactly what is happening in this pulley setup. I have looked closer at my function 9.8*500/2*cscx and realized that I could integrate this function to find the change in tension as the angle increases. This ends up being a number higher then it would take to simply lift this weight with a single rope directly up. This Equation is simply Mass*Gravity*Height, with a single rope pulling straight up to lift the mass. However, the system I am discussing has other forces that are impacting it's pull on the way down. I have found that this function can be integrated as 9.8*500kg(Any mass will do)*integral csc(theta) from pi/36 to pi/4. After doing many calculations, this combination seemed the most reasonable to obtain and yielded the highest work. The calculation from wolfram alpha clearly said that this value ends up being 11,024 joules. This is because the function that is integrated is taking into account distance, as the bounds themselves are what puts a certain value on this function. This function also follows the laws of trigonometry, closely, so i imagine it would be nearly impossible for me to break it. I have set up a situation where I mapped the sides of the triangle that this rope would create. I have measured the opposite side of the triangle and found that the distance is the square root of two, with a theoretical adjacent side of 1. So, in order to lift such a mass the height, I would need to place it on this device, I would calculate m*g*h. My h in this case would be square root of 2, for whatever unit I chose. Now I can calculate the work done to lift such a mass in order to place it in my system. It would now be 9.8*500kg*square root of 2(whatever units chosen) which ends up being 6929 joules. But if I was to set up generators in the pivotal points that the tension was connected to, I could utilize the work of both pulleys, each with half of the total work of 11,024. So each pivotal point would have the work of 5,512 joules. This would still be the work done by the weight itself, for each tension t1 and t2. I am neglecting some forces with such a system but I am also remember that this work would be done, even with friction. I also cannot break the laws of trigonometry which clearly says that I can integrate cscx to find the work done with respect to theta(the distance). I am more then happy to expand on this farther with pictures and of course am welcome to attacks on this theory because the math has backed me up to the point where I have been tempted to build such a device but not have had the resources or the money. I also see it pointless unless the math is screaming directly in my face and backing up my claims. I will attempt to attach any scratch work, although I would like to discuss this further with someone interested in this thought out theory. I have also considered more things then it may appear.

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Offline isodecryptor

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Re: Gravity seems to yield unequal results.
« Reply #1 on: February 22, 2014, 02:05:51 AM »
I also have messed up the numbers. With a theoretical adjacent side of a 1, my opposite would also be a 1, going to upper bound of 45 degrees. This works out better, because it would actually take less joules to lift. I also messed up the integration. It ends up being alot more. This is of course if I could get the system to have zero acceleration except for the very initial, in which the angle would start being calculated then.
« Last Edit: February 22, 2014, 07:57:05 AM by isodecryptor »

Offline isodecryptor

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Re: Gravity seems to yield unequal results.
« Reply #2 on: May 26, 2014, 11:54:54 PM »
I have added pictures to help explain some of the ideas. This integration is 9.8*500kg*integral cscx with bounds chosen of pi36 to pi 3. The integration ,integrates with distance being theta. Using similar triangles, i chose a shorter opp so it would take less energy to lift it back into the system with later introduced t3. It turns out that it does not matter what the actual size is, as long as it uses similar triangles and proportionality with the angle bounds chosen, so the function would be same no matter size opp or adjacent is. Seeing as how i must lift against gravity, i chose my opp to be shorter, as the work done from t3 is merely m*g*h, and height would represent my opp. of course i adjusted my other sides with proportionality of the already defined angles for the right triangle. I am only trying to perfect the model before considering any new forces.

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Re: Gravity seems to yield unequal results.
« Reply #2 on: May 26, 2014, 11:54:54 PM »
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Offline isodecryptor

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Re: Gravity seems to yield unequal results.
« Reply #3 on: June 20, 2014, 12:25:13 PM »
I have added to my idea. I am attaching a picture to try to help visualize what my idea is. I will have two solid poles that are close enough together so that when t1 and t2 have reached the bottom where the weight has fallen, it reaches the farthest possible angle, while still keeping the opposite or height of the pole the same. (Around 1 meter for analytical purposes.) Obviously, in order to achieve such an angle, the distances between the poles has to be reduced, if the intended end angle is to be large. So these poles would have attachments with bearings that hold a moveable spindle part, for the weights on the outside, and a spindle part that is shaped like a screw for the dropping weight on the inside ,which are all attached to the same axle. (Trying to utilize the changing force with theta.) I'm assuming that the less of a radius, the more force it would take to turn(but also could spin the outside spindle that has a larger radius, further). However, I am trying to convert force into distance. While my tension changes with theta, I would want my screw-spindle to utilize the larger force available with the smaller angle. The only way I can imagine moving the weights on the outside more is to have that larger force utilized in a torque ratio that changes with theta. That is where the "screw-spindle" seems to be of most use. It would simply wrap closer to the moment of inertia, as the angle becomes smaller( weight is higher up) and the radius gets larger from the moment, as the weight falls down. From my basic understanding of torque, the mechanical advantage comes from having to move further, (same work) but puts more force on the moment of inertia, or torque center particle. It seems to be exactly what my machine is attempting to do because the tension is changing. The work done is exactly the same with torque, but distance is actually converted into force(they seem proportional to each other) This machine is attempting to convert gravity into distance, as crazy as that sounds. Or changing force due to gravity and it's vertical angle theta,  into distance, with changing torque, accommodating changing tension with angle theta. I am not a master physicist but all of these ideas do seem logical from my mathematical and research background. They are implying things that are clearly stated impossible, but really seem to catch my attention with my limited understanding of the laws of physics and my minor in math. I must add that I missed one tension, which is not always a part of this closed system. It would be a rope, going straight up from the middle weight to pull this machine back into its original position, which I have calculated to be simply, MGH, work=mgh or potential. This rope would only be truly introduced to lift the weight back up one meter. (Which would seem to reset its original configuration without having to solve for the system in reverse, which obviously would be exactly equal to the work done down to lift back up. But I will not be using those ropes, because it would take less to lift it with a temporarily introduced rope to pull it back, straight up.

Offline isodecryptor

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Re: Gravity seems to yield unequal results.
« Reply #4 on: June 21, 2014, 04:17:16 AM »
another interesting assumption is that the magnitude of a can be thrown in for the height of the pole, to address differences in hpw much work wpuld need to be done in multiple setups, so work could look something like t1(@)=(a*m*g )/2*integral of,csc(@) from a to b degrees) Because theta is the x or distance for this function, however the magnitude of the opp could vary. I have been using a standard measure of 1 meter, to represent my unit circle in the proportionality is is naturally configured in. But A could vary, so this would seem to adjust the work needed by t1 and t2 for any particular setup.

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Re: Gravity seems to yield unequal results.
« Reply #4 on: June 21, 2014, 04:17:16 AM »
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Offline isodecryptor

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Re: Gravity seems to yield unequal results.
« Reply #5 on: June 22, 2014, 06:50:03 AM »
Im going to add a rough estimate of the frictional losses, because friction is a real part of our world. so workt1(@)=(a*m*g)/2*integral cscx from 5 to 85 degrees -(uk*a*m*g)/2*integral cscx from 5 to 85 degrees. uk is kinetic friction, a =
height of the pole. a pretty high kinetic friction coefficient, for ball bearing, .3 (much higher than it would be by far) work t1(@)= ((1*500*9.8)/2*integral cscx 5 to 85 - (.3*1*9.8*500)/2*integral cscx from 5 to 85) Because friction would increase and decrease with theta as well. This turns out to be workt1(@)=7457.6-heat loss to friction(2237.28) joules. That would be roughly 5220.6 joules of work for t1 and 5220.6 joules of work done was not lost as heat due to friction. So it could do 10,441.2 joules of productive work.

Offline isodecryptor

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Re: Gravity seems to yield unequal results.
« Reply #6 on: June 25, 2014, 01:46:30 PM »
I built a poorly constructed model out of old fishing poles. I must say the results were interesting and hard to interpret. I played with the amounts of weight on the outsides. What immediately caught my attention is that the middle weight could drop no more then the weights would on the outside would rise. So it seemed that this would make calculating the work done easier. What is confusing is that the middle weight could in fact pull the outside weights up, which did actually weigh more. And of course these weights rose equally as much as the middle weight fell. The middle weight has to have twice the amount of line to fall the same distance as either of the outside weights. So there is no mistaking how much potential from the middle wieght is lost, Compared to the new potential of the weights on the outside, who's sum of weight did equal more than the inner weight that is falling. Different outer weights m2+m3 were always more than m1. As i would add more weight to m2 and m3, the rise would decrease, but the middle weight would never fall further than the outside weights would rise. The confusing part is the middle weight would fall short of falling all the way, so some of its potential remained in the system at equal librium. So it has the feeling and look of overunity, but i am having trouble being sure to calculate energy in, opposed to out. there were times when i was almost certain it had overunity abilities, but will of course remain skeptic, as i do have a limited understanding of physics. Otherwise, it has some mechanical advantages that could be used to do work. I have never seen anything quite like it and am baffled at the results and how to interpret them, nevertheless. I would say it is at least invention worthy. especially if you needed to do some heavy lifting.

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Re: Gravity seems to yield unequal results.
« Reply #6 on: June 25, 2014, 01:46:30 PM »
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Offline isodecryptor

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Re: Gravity seems to yield unequal results.
« Reply #7 on: June 27, 2014, 06:10:38 AM »
I am going to add a detail writing of my findings. I must also add that my machine was thrown together just to see if it behaved in the ways that i suspected, mathematically. The keeping of the symmetry of the function cscx is highly destitute to the expected results. The less the mechanical configuration represents cscx, the less work it does by a very significant amount. My middle weight was initially a fishing weight that had a hook embedded in it. This allows the weight to free  shift on the fishing line, which allowed physics to adjust the weight into the middle, helping to reserve its closer symmetrical representation for cscx. I did not realize how important this truly was until i changed my weights, using a regular solenoid shaped weight for the middle. This did not slide as well, and also altered the connections of the two tensions holding it up, and SEVERELY affected the amount it could do on the way down. Once i realized this I put a proper hooking device to it that was cut to have a negligible impact on the mass of m1. I have a strong suspicion that the cscx asymptotic behaivior is having some very interesting affects on my machine.Even enough for a possible overunity affect. Please remain ooen minded, as i share my personal experiments, free of charge.
 :-t

Offline isodecryptor

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Re: Gravity seems to yield unequal results.
« Reply #8 on: June 27, 2014, 06:24:46 AM »
Here is a more detailed list of my configurations and findings.

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Re: Gravity seems to yield unequal results.
« Reply #8 on: June 27, 2014, 06:24:46 AM »
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Offline isodecryptor

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Re: Gravity seems to yield unequal results.
« Reply #9 on: June 28, 2014, 06:14:13 PM »
Another equally far fetched idea is to have two of the structures facing each other with an equally balanced lever under the two middle weights.

Offline isodecryptor

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Re: Gravity seems to yield unequal results.
« Reply #10 on: June 28, 2014, 06:16:53 PM »
Here is an idea that i cannot prove. But interesting.

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Re: Gravity seems to yield unequal results.
« Reply #10 on: June 28, 2014, 06:16:53 PM »
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Offline isodecryptor

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Re: Gravity seems to yield unequal results.
« Reply #11 on: June 28, 2014, 09:54:45 PM »
The reason i would like to see this configuration is because it would take alot less work to lift the middle weights back up, once they have the seemingly additional potential of m2 and m3, which would assist the middle weight to go back up. I know it would take significantly less work to return the middle weights back to the top, now that there is a seemingly surplus in potential from m2 and m3. So the falling weight would be less impacted to push it back up in a vertical direction, especially since cscx has the asymptotic behaviors with the changing angle. Remember that this would be considered a vertical lift. I will try to find an integral that would suit the work needed to lift the middle weight up vertically, with the assisting vertical vector components of t1 and t2 from m2 and m3. If its equal, than the function is just confusing. If nothing else, it may make a cool see saw. Or a carnival ride :).

Offline isodecryptor

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Re: Gravity seems to yield unequal results.
« Reply #12 on: June 28, 2014, 10:00:11 PM »
and please, if anyone is good with physics simulation software, i would love to collaborate on a model of this.

Offline isodecryptor

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Re: Gravity seems to yield unequal results.
« Reply #13 on: June 29, 2014, 01:17:00 PM »
A cool result i noticed with varying outside weights is the, oh shit, we weigh more then him affect. The middle weight would fall so far, lifting so much mass that it would pause for a moment, then readjust the middle weight back up to a tension equal in magnitude with respect to theta. This only took place when the outside weights sum was more than 1.5 times the middle weight. I am aware of elasticity, but this looke slightly different. It was an abnormally long pause that took a while longer to reconfigure than expected. Almost like the outside weights themselves were the main force that was spinning it bqck up, to reach equalibrium. But it seems as though equilibrium would have allowed no movement at all. In almost all other work, torque, tensions ,relationships i have observed. And if there was a difference in weight, it would appear as a mechanical advantage trade off. i move more less distance etc etc. Not, i move more the same distance, then confuse myself as to what just happened.


Offline isodecryptor

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Re: Gravity seems to yield unequal results.
« Reply #14 on: June 30, 2014, 12:07:21 PM »
More speculations.

 

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