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Here is how to do it

Anything to do with the electric or hybrid world

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gruntguru
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Re: Here is how to do it

Post by gruntguru » Mon Jul 15, 2019 12:17 am

exhaustgases wrote:
Fri Jul 12, 2019 6:25 pm
Ant180 wrote:
Fri Jul 12, 2019 10:43 am
Wow. Just wow.
I don’t know how this can seriously be an argument on this forum, unless perhaps EG’s account has been hacked.
The solution is simple guys. Let’s leave EG to build the device and become a billionaire by making energy, while we pay market rates for said energy.
EG, you are a genius, and we are all morons!
I did the experiments years ago. It really is not a difficult thing to do. More windings use less energy to make more mag strength. Its simple ohms law stuff. Back in the day some thought Tesla was an idiot. For that matter Einstein was poo'ed on as well. No I'm not close to any of them.
Just wind some coils and check it out.
Everyone here (except yourself) knows what the results of "winding some coils" will be. Its all covered in the text books.

What YOU need to do is build one of these fantasy machines that can produce motion with any external energy input or depletion of any internal energy source. None of the devices you have linked in your posts actually work - really!!!
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Re: Here is how to do it

Post by GRTfast » Tue Jul 16, 2019 12:11 pm

gruntguru wrote:
Mon Jul 15, 2019 12:17 am
exhaustgases wrote:
Fri Jul 12, 2019 6:25 pm
Ant180 wrote:
Fri Jul 12, 2019 10:43 am
Wow. Just wow.
I don’t know how this can seriously be an argument on this forum, unless perhaps EG’s account has been hacked.
The solution is simple guys. Let’s leave EG to build the device and become a billionaire by making energy, while we pay market rates for said energy.
EG, you are a genius, and we are all morons!
I did the experiments years ago. It really is not a difficult thing to do. More windings use less energy to make more mag strength. Its simple ohms law stuff. Back in the day some thought Tesla was an idiot. For that matter Einstein was poo'ed on as well. No I'm not close to any of them.
Just wind some coils and check it out.
Everyone here (except yourself) knows what the results of "winding some coils" will be. Its all covered in the text books.

What YOU need to do is build one of these fantasy machines that can produce motion with any external energy input or depletion of any internal energy source. None of the devices you have linked in your posts actually work - really!!!
Top
...but he can't build a simple proof of concept because he doesn't have the money. :lol:

exhaustgases
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Re: Here is how to do it

Post by exhaustgases » Tue Jul 16, 2019 10:36 pm

Just take some copper mag wire and wrap it around a 3/8 bolt about 2.5 long, use maybe 40 turns, measure the ohms. Apply 1.5 volts to the coil and see how strong it is.
Then make another with 300 turns of the same wire and do the same. Tell me what you find out.
The less resistant coil of 40 turns will need more power to work, were as the other one uses less power and is stronger, proof of concept. Don't have the wire handy to measure the diameter / gauge.

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Re: Here is how to do it

Post by GRTfast » Wed Jul 17, 2019 7:07 am

exhaustgases wrote:
Tue Jul 16, 2019 10:36 pm
Just take some copper mag wire and wrap it around a 3/8 bolt about 2.5 long, use maybe 40 turns, measure the ohms. Apply 1.5 volts to the coil and see how strong it is.
Then make another with 300 turns of the same wire and do the same. Tell me what you find out.
The less resistant coil of 40 turns will need more power to work, were as the other one uses less power and is stronger, proof of concept. Don't have the wire handy to measure the diameter / gauge.
How does this in any way prove that you can make a motor that only uses permanent magnets as an energy source, or that you can make a motor that outputs more energy than it takes to run it? Be specific.

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Re: Here is how to do it

Post by gruntguru » Wed Jul 17, 2019 11:58 pm

exhaustgases wrote:
Tue Jul 16, 2019 10:36 pm
Just take some copper mag wire and wrap it around a 3/8 bolt about 2.5 long, use maybe 40 turns, measure the ohms. Apply 1.5 volts to the coil and see how strong it is.
Then make another with 300 turns of the same wire and do the same. Tell me what you find out.
The less resistant coil of 40 turns will need more power to work, were as the other one uses less power and is stronger, proof of concept. Don't have the wire handy to measure the diameter / gauge.
Yep - the 300 turn electromagnet has a stronger magnetic field and the electrical power required to maintain that field is lower.

Next, calculate the time taken to build the magnetic field (longer for the 300 turn case) and how much electrical energy was used to build that field (this is what has to happen if you want to use the electromagnet in a motor - you have to switch the field on and off). Hmmm - more time and more energy required to charge the 300 turn E.M. Perhaps you can't get something for nothing.

Yes exhaustgases, I have done all these experiments as part of my study. It is YOU who needs do the experiment - build the perpetual motion machine and find that it doesn't work. Those of us who understand the theory are not going to waste our time trying to disprove the electrical machine fundamentals that have been refined over hundreds of years by smarter men and women than you and I.

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Re: Here is how to do it

Post by exhaustgases » Thu Jul 18, 2019 3:44 pm

Grunt, now capture that counter emf, and use it. At least we proved we can get more mag force for less, thank you.
Time ? The device doesn't have to run at thousands of RPMs.

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Re: Here is how to do it

Post by GRTfast » Thu Jul 18, 2019 6:37 pm

exhaustgases wrote:
Thu Jul 18, 2019 3:44 pm
At least we proved we can get more mag force for less, thank you.
Formula for solenoid force:

F = (n*i)2*magnetic constant*a/(2*g^2)

Where:
F = Force, i = Current, g = Length of the gap between the solenoid and a piece of metal, a = Area n = Number of turns, Magnetic constant = 4*Pi*10^-7.

What happens to the resistance of a wire when you make it longer? it increases linearly with respect to the length. What happens to the current when the resistance increases? It decreases linearly with respect to resistance.

This means that the current decreases linearly with respect to the length of the wire.

What happens to the number of turns as the wire length is increased? It increases linearly with respect to the length of the wire.

So, as we increase the number of turns, the current decreases at the same exact rate as the number of turns increases. In your suggested experiment, n*i remains constant. If n*i remains constant, and all the other values on the right side of the equation are constant, what does that say about the force? Hopefully you can put that together for yourself.

Now, if you add turns and you keep the current constant, the force will increase....BUT what do you have to do to keep the current constant? increase the voltage. It takes more power to make the solenoid apply more force.

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Re: Here is how to do it

Post by exhaustgases » Thu Jul 18, 2019 8:19 pm

Just worry about using a set voltage and tell me what happens to the supplied power, when increasing the windings.
Your too busy trying to make something simple complicated.

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Re: Here is how to do it

Post by GRTfast » Thu Jul 18, 2019 9:15 pm

exhaustgases wrote:
Thu Jul 18, 2019 8:19 pm
Just worry about using a set voltage and tell me what happens to the supplied power, when increasing the windings.
Your too busy trying to make something simple complicated.
You need to consider the internal resistance of the power source, as it is part of the circuit. You also need to consider magnetic saturation, and the material properties of the wire limiting the temperature it can tolerate. This is essentially an optimization problem. For a given set of parameters, the coil dimensions can be found such that the best efficiency is achieved....and it will never ever be over 100%.

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Re: Here is how to do it

Post by gruntguru » Thu Jul 18, 2019 9:57 pm

exhaustgases wrote:
Thu Jul 18, 2019 3:44 pm
Grunt, now capture that counter emf, and use it. At least we proved we can get more mag force for less, thank you.
Time ? The device doesn't have to run at thousands of RPMs.
The key thing is that to use an EM as a motor, you have to CHANGE the magnetic field strength. To change the magnetic field strength you have to supply energy. The energy you get out of your motor will never be as much as the energy you have to put in - period. Stop pissing in the wind - build it - prove us wrong.

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Re: Here is how to do it

Post by Nefario » Thu Jul 18, 2019 10:10 pm

exhaustgases wrote:
Thu Jul 18, 2019 8:19 pm
Just worry about using a set voltage and tell me what happens to the supplied power, when increasing the windings.
Very simply - I didn't study much about coil design - Electrical power = V x I (volts x amps)

Ohms Law: V = IR (amps x resistance), I = V/R

Say R-40 turns = "40", R-300 = "300"

Power = V**2/R --> Power40 = V**2/40, Power300 = V**2/300, At a set voltage V the larger coil simply acts as a larger resistor and LIMITS the amounts of power than can flow through the winding. The magnet may be more efficient - or not - but magnetic energy out can't be greater than electrical energy in - this is the first s̶u̶g̶g̶e̶s̶t̶i̶o̶n̶ law of thermodynamics.
Dave

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Re: Here is how to do it

Post by exhaustgases » Sat Jul 20, 2019 1:32 pm

Nefario wrote:
Thu Jul 18, 2019 10:10 pm
exhaustgases wrote:
Thu Jul 18, 2019 8:19 pm
Just worry about using a set voltage and tell me what happens to the supplied power, when increasing the windings.
Very simply - I didn't study much about coil design - Electrical power = V x I (volts x amps)

Ohms Law: V = IR (amps x resistance), I = V/R

Say R-40 turns = "40", R-300 = "300"

Power = V**2/R --> Power40 = V**2/40, Power300 = V**2/300, At a set voltage V the larger coil simply acts as a larger resistor and LIMITS the amounts of power than can flow through the winding. The magnet may be more efficient - or not - but magnetic energy out can't be greater than electrical energy in - this is the first s̶u̶g̶g̶e̶s̶t̶i̶o̶n̶ law of thermodynamics.
I don't have the wire at this time. Just for example a wire with .0842 ohms per foot, say the first coil is 2.5 feet long of wire and the second one is 20 feet long. The 2.5 foot wire is .2105 ohms. The 20 foot wire is 1.684 ohms. The power source is 1.5 volt battery. Simple ohms law stuff the 2.5 foot wire takes 7.12 amps, and the 20 foot wire takes .890 amps. The power equation is P=IE yeah pie.....So the 2.5 foot wire on the small winding coil takes 10.68 watts, and the 20 foot wire on the large coil takes 1.335 watts. So by this we see the big and more powerful magnetic force coil takes way less power to operate than the small less powerful one does. And I tested this over 30 years ago.

Nefario ? So what are you saying and accomplishing with the above? Just try what I just posted. The mag force is greater for the larger coil. Just see what each coil will pick up. Everyone wants me to do it I have over 30 years ago. Now its your turn to do it.

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Re: Here is how to do it

Post by exhaustgases » Sun Jul 21, 2019 5:05 pm

exhaustgases wrote:
Sat Jul 20, 2019 1:32 pm
Nefario wrote:
Thu Jul 18, 2019 10:10 pm
exhaustgases wrote:
Thu Jul 18, 2019 8:19 pm
Just worry about using a set voltage and tell me what happens to the supplied power, when increasing the windings.
Very simply - I didn't study much about coil design - Electrical power = V x I (volts x amps)

Ohms Law: V = IR (amps x resistance), I = V/R

Say R-40 turns = "40", R-300 = "300"

Power = V**2/R --> Power40 = V**2/40, Power300 = V**2/300, At a set voltage V the larger coil simply acts as a larger resistor and LIMITS the amounts of power than can flow through the winding. The magnet may be more efficient - or not - but magnetic energy out can't be greater than electrical energy in - this is the first s̶u̶g̶g̶e̶s̶t̶i̶o̶n̶ law of thermodynamics.
I don't have the wire at this time. Just for example a wire with .0842 ohms per foot, say the first coil is 2.5 feet long of wire and the second one is 20 feet long. The 2.5 foot wire is .2105 ohms. The 20 foot wire is 1.684 ohms. The power source is 1.5 volt battery. Simple ohms law stuff the 2.5 foot wire takes 7.12 amps, and the 20 foot wire takes .890 amps. The power equation is P=IE yeah pie.....So the 2.5 foot wire on the small winding coil takes 10.68 watts, and the 20 foot wire on the large coil takes 1.335 watts. So by this we see the big and more powerful magnetic force coil takes way less power to operate than the small less powerful one does. And I tested this over 30 years ago.

Nefario ? So what are you saying and accomplishing with the above? Just try what I just posted. The mag force is greater for the larger coil. Just see what each coil will pick up. Everyone wants me to do it I have over 30 years ago. Now its your turn to do it.
The silence is deafening.

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Re: Here is how to do it

Post by GRTfast » Sun Jul 21, 2019 6:18 pm

exhaustgases wrote:
Sun Jul 21, 2019 5:05 pm
The silence is deafening.
The force is not stronger for a coil with more turns (all else equal) at the some voltage as a coil with less turns. The force is constant in an ideal case, but in reality it decreases as you add more turns. There is an optimum coil geometry for a given constant voltage power source.

http://citeseerx.ist.psu.edu/viewdoc/do ... 1&type=pdf

What is your point again?

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Re: Here is how to do it

Post by Nefario » Sun Jul 21, 2019 9:13 pm

exhaustgases wrote:
Sat Jul 20, 2019 1:32 pm
....So the 2.5 foot wire on the small winding coil takes 10.68 watts, and the 20 foot wire on the large coil takes 1.335 watts. So by this we see the big and more powerful magnetic force coil takes way less power to operate than the small less powerful one does. And I tested this over 30 years ago.
"The silence..." was taking a weekend break....

The 20-foot wire provides enough windings to create a stronger magnetic field. The shorter wire make fewer windings, generates fewer lines of force, and is converting more of the electric power to heat.
Dave

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