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Author Topic: Spring Coupling  (Read 14749 times)

Offline vineet_kiran

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Re: Spring Coupling
« Reply #15 on: July 26, 2014, 03:03:44 PM »
These are very basic concepts that you can check out for yourself:

The product of force and distance moved on one side of a lever is, absent losses, the same as the product of force and distance on the other side of a lever. 

Coupling through different size pulleys, gears, etc, is just a rotary version of a lever.

I have already proven mathematically that your concept does not work as you claim.

 
I agree with that.

a) In a lever force gets magnified and distance gets reduced.

   What happens if in a lever force gets magnified and distance is not reduced?  - which is practically
   not possible

b) In gears and pulleys (rotary version of lever) torque gets magnified and speed gets reduced.
   What happens if in gears and pulleys torque gets magnified and speed is not reduced?
 
   Which is practically possible using a spring coupling.

   It is also possible by another method.  In the following link PDF file,  please see 'Mechanical analogy'
   
   http://www.overunity.com/10774/over-unity-by-reaction-helping-action/msg287484/#msg287484
 
 
   In your 'mathematical analysis'   you have considered only E (in) and E (out)  but not torque and speed.
 
 
 
 
 
 
 

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Re: Spring Coupling
« Reply #15 on: July 26, 2014, 03:03:44 PM »

Offline MarkE

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Re: Spring Coupling
« Reply #16 on: July 26, 2014, 03:25:35 PM »
You postulate conditions that you cannot show to exist.  That is called fantasy.

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Offline vineet_kiran

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Re: Spring Coupling
« Reply #17 on: July 26, 2014, 03:59:03 PM »
You postulate conditions that you cannot show to exist.  That is called fantasy.

 
You conclude without verifying.  That is called ignorance.
 
 
I am not formulating any postulate.    I have posted the experiment in "Half Baked Ideas"  for discussion.
 
 
 
 

Offline MarkE

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Re: Spring Coupling
« Reply #18 on: July 26, 2014, 04:52:39 PM »
Your postulate relies on the false notion that the average tension within the spring is not constant.   Basic physics teaches us that is false.  You are free to try and show that such a thing is possible.  Simply declaring fantastical ideas does not either make them true or worthy of debate.

Your assertion:

"When motor shaft is rotated, it rotates the generator shaft applying force at the end of larger diameter producing greater torque at the centre of generator shaft but the reactive force of generator flange acts on outer edge of motor flange since spring is flexible and cannot offer reaction force.  Hence you will get mechanical advantage depending on ratio of diameters of flanges with same speed."

Is wrong on multiple counts.  The spring conveys the same torque to the big end as the small end.  The tangential forces vary by radius.  The torque does not.


Offline vineet_kiran

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Re: Spring Coupling
« Reply #19 on: July 26, 2014, 06:03:44 PM »

Your postulate relies on the false notion that the average tension within the spring is not constant.   Basic physics teaches us that is false.  You are free to try and show that such a thing is possible. 
 

A huge rotating wheel will have tremendous moment of inertia.  Hence once you set a huge wheel in rotation,  force to be applied on its outer edge to keep it rotating will be negligible. The tension developed in spring will be corresponding to this force only.  So, tension developed in spring and energy consumed for tensioning the spring are also negligible.  This will be taken care of by mechanical advantage.
 

Simply declaring fantastical ideas does not either make them true or worthy of debate.


 
What for this forum is there?   What for "half baked ideas"?
 
 

The spring conveys the same torque to the big end as the small end.  The tangential forces vary by radius.  The torque does not.


That happens only in case of rigid coupling.  If you tie a meter length rope to a pillar and apply tangential force at the end of the rope,  will the tangential force depend on the radius?  Whatever force you apply at the edge of the rope,  same force appears at the outer edge of the pillar. 
 
For rigid coupling  F(tangential) generator is less than F (tangential) motor depending on radius.

For spring coupling F (tangential) generator  =  F (tangential) motor,  doesnot depend on radius.   Hence torque will be magnified.
 
 

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Re: Spring Coupling
« Reply #19 on: July 26, 2014, 06:03:44 PM »
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Offline MarkE

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Re: Spring Coupling
« Reply #20 on: July 26, 2014, 07:26:08 PM »

A huge rotating wheel will have tremendous moment of inertia.  Hence once you set a huge wheel in rotation,  force to be applied on its outer edge to keep it rotating will be negligible.
You are conflating temporarily consuming stored kinetic energy with torque transfer.
Quote

The tension developed in spring will be corresponding to this force only.  So, tension developed in spring and energy consumed for tensioning the spring are also negligible.  This will be taken care of by mechanical advantage.

Wrong.  The tension in the spring is a result of the reflected load torque.  Over time the power contributed by a finite energy store such as the maximum energy that the spring can store or the flywheel can store approaches zero. 
Quote

 
What for this forum is there?   What for "half baked ideas"?
The idea is dead on arrival because it relies on non-physical concepts.
Quote

 
That happens only in case of rigid coupling.
Again that is wrong.  The compliance of the coupling sets a maximum variation, and combined with the inertia sets a time constant.  Averaged over many time constants, the torque is the same as with a rigid coupling.
Quote

 If you tie a meter length rope to a pillar and apply tangential force at the end of the rope,  will the tangential force depend on the radius?  Whatever force you apply at the edge of the rope,  same force appears at the outer edge of the pillar. 

I see you appear to be very confused about where the forces originate.   If you apply a linear tangential force out at some moment arm as in your pillar example then the torque depends on the radius.  A rotating device such as a motor develops a torque, not a linear tangential force.  A generator reflects a torque, not a linear tangential force.
Quote

 
For rigid coupling  F(tangential) generator is less than F (tangential) motor depending on radius.
Both the motor and the generator operate on torque.  The spring transfers torque.  The tangential forces depend on the radii.
Quote

For spring coupling F (tangential) generator  =  F (tangential) motor,  doesnot depend on radius.   Hence torque will be magnified.
No dude,  see as I have explained.  If I have an 1800 rpm 1 hp motor, that motor turns 188.5r/s and produces a torque of 3.96 N*m.  Tangential force, not the torque depends on the radius.  Ditto a generator that produces 1hp*efficiency output at 1800 rpm.

Offline webby1

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Re: Spring Coupling
« Reply #21 on: July 26, 2014, 07:40:18 PM »
If the thought is that if I pull on one end of a clock spring with 1lb of force then the other end of the clock spring needs to hold against that force, and if the length of arm connecting the two ends are not the same then the resulting torque on the output side will be higher,, sound enough except for the change in direction, or angle, of force as it is transferred through the spring,, the ends, in this case, end up pulling in towards each other as well as a tangential force against the ends of the arms.

I say this because I actually built this about a year ago,, and noticed that the end connections would rotate as force was applied pointing towards the true direction of applied force.

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Re: Spring Coupling
« Reply #21 on: July 26, 2014, 07:40:18 PM »
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Offline MarkE

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Re: Spring Coupling
« Reply #22 on: July 26, 2014, 07:50:12 PM »
It needs to hold against the torque.  A rigid coupling is nothing more than a spring with a very high K.

Offline vineet_kiran

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Re: Spring Coupling
« Reply #23 on: July 27, 2014, 06:55:39 AM »
If the thought is that if I pull on one end of a clock spring with 1lb of force then the other end of the clock spring needs to hold against that force, and if the length of arm connecting the two ends are not the same then the resulting torque on the output side will be higher,, sound enough except for the change in direction, or angle, of force as it is transferred through the spring,, the ends, in this case, end up pulling in towards each other as well as a tangential force against the ends of the arms.

I say this because I actually built this about a year ago,, and noticed that the end connections would rotate as force was applied pointing towards the true direction of applied force.

 
You are exactly right.  The force appearing on generator flange depends on direction or angle of application of force from the spring.  To overcome this problem you may have to use several springs one above the other with their ends terminating at equidistances on respective flanges.
 
For example if you use four springs, their ends should terminate at 90 degrees apart on respective flanges. You know about this because you have conducted the experiment.  It is difficult argue with people who simply talk without practical knowledge.
 
 
 

No dude,  see as I have explained.  If I have an 1800 rpm 1 hp motor, that motor turns 188.5r/s and produces a torque of 3.96 N*m.  Tangential force, not the torque depends on the radius.  Ditto a generator that produces 1hp*efficiency output at 1800 rpm.
   

 
The torque developed by a 1 HP motor depends on its internal characteristics such as internal diameter of the rotor, strength of electromagnetic forces etc.   The torque you get at the motor shaft is a result of these internal characterisics.

But if you fix a lengthy lever arm to the motor shaft,  even a little child can hold this lever arm and prevent 1 HP motor shaft from rotating.  If you tie a lengthy rope instead of a lever arm,  even a adult cannot prevent motor shaft from rotating because force doesnot get magnified through rope since it is not rigid.

Through your postings I can make out that you are a superficial crank.  Too much book reading without practical knowledge makes a superficial crank.
 
Take examples of Farday, Newton, Edison  etc.,  I think none of them had academic carreer.
 
 
 

Offline MarkE

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Re: Spring Coupling
« Reply #24 on: July 27, 2014, 10:13:10 AM »
The torque developed by a 1 HP motor depends on its internal characteristics such as internal diameter of the rotor, strength of electromagnetic forces etc.   The torque you get at the motor shaft is a result of these internal characterisics.
Those characteristics are how the motor gets its 1hp rating.
Quote

But if you fix a lengthy lever arm to the motor shaft,  even a little child can hold this lever arm and prevent 1 HP motor shaft from rotating.  If you tie a lengthy rope instead of a lever arm,  even a adult cannot prevent motor shaft from rotating because force doesnot get magnified through rope since it is not rigid.
And in that you have just confirmed that the tangential force is as I have been telling you the motor torque divided by the length of the moment arm.  The torque into the spring does not depend upon the flange plate or spring diameters.  You have now eviscerated your own postulate.
Quote

Through your postings I can make out that you are a superficial crank.  Too much book reading without practical knowledge makes a superficial crank.
 
Take examples of Farday, Newton, Edison  etc.,  I think none of them had academic carreer.
Ad homs on top of destroying your own argument don't speak well for you.

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Re: Spring Coupling
« Reply #24 on: July 27, 2014, 10:13:10 AM »
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Offline webby1

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Re: Spring Coupling
« Reply #25 on: July 27, 2014, 05:55:27 PM »
I think an overly simplified view might help.

One of the attributes of the spring is its length.  Like a string, the length of the spring does not change and so if one end is pulled 3 feet the other end will also move 3 feet.  This does not matter if the pull is in a straight line or if it is in a circular path.

If the inside end of the spring is on a 1 foot radius and the outside end of the spring is on a 5 foot radius,, then what?

If I rotate the inside of the spring so that I have moved that end 3 feet then the outside end of the spring can only move 3 feet, but to keep up rotationally the outside end of the spring needs to move 15 feet.

If I rotate the inside of the spring 4 revolutions while holding the outside of the spring stationary, then I release the outside end of the spring while I am making the 5th revolution of the inside end then the outside end will make 1 revolution while the inside end is making 1 revolution, but the spring that has been wound up will unwind to make up the extra distance of travel needed by the outside end.

The extra distance of motion with the required force can be stored within the "spring" attribute of the spring, but this is a temporary holding space for input work that has been performed but not used.

Offline MarkE

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Re: Spring Coupling
« Reply #26 on: July 27, 2014, 09:50:41 PM »
Tom, sure you can do those things.  However, over time you can't either keep winding up the spring indefinitely, nor can you unwind it indefinitely.  Over time the average amount of wind-up or relaxation on a per rotation basis tends towards zero.  That means that on average the amount of power that the spring stores or releases tends towards zero.  On average the spring just conveys power from the motor to the load (generator).  The input power is the motor torque multiplied by the shaft speed.  That same power, ignoring losses, conveys from one end of the spring to the other. 


Offline ARMCORTEX

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Offline webby1

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Re: Spring Coupling
« Reply #28 on: July 28, 2014, 01:06:06 AM »
Tom, sure you can do those things.  However, over time you can't either keep winding up the spring indefinitely, nor can you unwind it indefinitely.  Over time the average amount of wind-up or relaxation on a per rotation basis tends towards zero.  That means that on average the amount of power that the spring stores or releases tends towards zero.  On average the spring just conveys power from the motor to the load (generator).  The input power is the motor torque multiplied by the shaft speed.  That same power, ignoring losses, conveys from one end of the spring to the other.

Mark,

What I am saying is that the work in is the same as the work out, but I am putting it in another form.  That is excluding losses due to heat and friction.
 

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Offline ARMCORTEX

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Re: Spring Coupling
« Reply #29 on: July 28, 2014, 01:21:20 AM »
.

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Re: Spring Coupling
« Reply #29 on: July 28, 2014, 01:21:20 AM »

 

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