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Author Topic: Where i'm at 1+1=3  (Read 14633 times)
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  Was looking into getting the bearings for the printed gears and noticed that they are the very same bearings that is used for inline skates. Have a box full, talk about off the shelf parts. Good choice Brad
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author=bistander link=topic=4603.msg111101#msg111101 date=1711602372


Quote
I disagree. Appears the graphs show a linear force (actually grams, which are used as a substitute for force) vs linear distance (mm).
Gram- Force is a valid force measurement unit.
The graphs show a rotational force being applied to the rotor of the motor.\

Quote
"torque over distance", it is not indicating energy or power.

Energy is the ability to do work. Power is the work done over time.
I said time is irrelevant, as i never claimed any power outputs were being calculated.
What was calculated was a gain in torque over distance between the two motors, where the two motors use the same rotor, same electromagnet, and same P/in
The graphs show a torque curve over a distance.

Quote
And a graph of "output torque v P/in" does not relate to any power output.

I never said anything about power output, as that would have to include time, mass, and acceleration, which is not what the graphs are indicating.

Quote
Time is relevant. You need rotational velocity to deduce power (and energy) from torque.

Time is not relevant, as we are not calculating power. Eg,
We are measuring the difference in torque over distance.
The graphs clearly show the applied torque over distance-work being done on the rotor over that short distance.

Quote
Recall my specific question:"The graphs tell me nothing about energy or work. That's the question I have, you say they do. How?"

Once again, the graphs show an applied torque over a distance.
I think you are looking at the graphs in a static manor, where they should be viewed as showing the rotor in motion over that distance. That is work being done.
Energy is the total amount of work done, and power is how fast you can do it.
As we are not calculating P/out, then time is irrelevant.
The total amount of work done is between the start of the torque curve, to the end of the torque curve.
The distance is between the start and end of the torque curve.

Quote
Side loading (or axial thrust) does not substract from output power directly, just from extra bearing friction loss.

It absolutely subtracts from total output rotational power.
The graphs prove that beyond doubt.
Those involved in that experiment will confirm this.
A higher torque applied over a distance directly increases rotational power output.
With my motor, side loading on the bearings is reduced by 75%+

Quote
Only the force in the direction of motion or displacement enters into the work formula.

That is exactly what the graphs are showing.
Force over distance in the direction of rotation.

Quote
Sorry, I didn't realize there were two different projects. I don't want to distract. Why don't we continue after you complete your plan? I really want to see you reveal this device.
bi

The graphs we are discussing, are of the motor which is the 1+1=3.
So the subject matter we are discussing is directly related to the graphs i posted.

The all magnet motor also being discussed here, is a different motor and subject.
So we are both discussing the same motor--the torque force motor.

Brad


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  Was looking into getting the bearings for the printed gears and noticed that they are the very same bearings that is used for inline skates. Have a box full, talk about off the shelf parts. Good choice Brad
thay

That is exactly why i use that size-they are common.
The same bearings are used in the 12/24 volt scooter motors.


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Thanks for the reply.
You mix the linear and rotational mechanical parameters and then power to energy comparison omitting time.

Power is volts * amps, held equal for the two graphs.

Power is (radians /second) * torque.

You say.
"My graphs show torque over a distance in a rotating system."
That would be your graphs show torque vs angle (or radians). And that is not energy.

The graphs tell me nothing about energy or work. That's the question I have, you say they do. How?

And I totally missed your explanation of "wasted torque". Is that side loading or radial thrust? How does that enter into energy besides bearing friction?

Like I said, it is difficult for me to follow your method. Perhaps I should have just waited to see where it leads. Which I am content to do.

Regards,
bi

Hope the diagram below makes things a little clearer.

Brad


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Hope the diagram below makes things a little clearer.

Brad

Hi Brad,
Yes, the diagram is helpful. Thanks.

Text in CAPS is copied and pasted from your diagram.

GRAPHS SHOW TORQUE OVER DISTANCE MEASURED BETWEEN POINT A AND B OF THE ROTATING MASS 

Actually the graphs show the linear force (in grams-force) vs linear distance (in mm). This is not the same as torque of the rotating mass.

THIS RESULTS IN A HIGHER ROTATIONAL OUTPUT POWER VALUE FOR THE SAME POWER INPUT TO THE ELECTROMAGNET.

(my underline)
So you do use the graph to make a deduction of rotational power based on torque. This brings time into the picture.

And I disagree with your work calculations due to omission of cos θ.

That's all.
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I took the time to find the graphs from the tests carried out in the 1+1=3 topic.
As you can see, the torque over distance increase in the modified motor, for the same input power, is greatly different.
Rotational force over distance is work being done.

Brad

Torque is to rotational motion what force is to translational motion. It has nothing to do with energy.

At each point, take the force, multiply it by the distance travelled between 2 points, with points close enough together that the force can always be considered to be constant and tangent to the line between two points. Then add all these products, calculated along a closed path, to give the energy, and only then can we say whether 1+1=3.



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Torque is to rotational motion what force is to translational motion. It has nothing to do with energy.
<snip>


Torque, or force, without motion has nothing to do with energy. But when you associate it with motion, as you did in your quote, then it, torque or force, has much to do with work, and energy.
My point of view.
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Torque, or force, without motion has nothing to do with energy. But when you associate it with motion, as you did in your quote, then it, torque or force, has much to do with work, and energy.
My point of view.
bi

That's exactly what I'm saying  C.C, except that I've also explained how to make this "association".

And unfortunately that's not what's been done so far.


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Believing in something false doesn't make it true.

Okay guys look at this thread, watch the video and read all of it and then explain why you think it can't work.

https://overunitymachines.com/index.php/topic,61.0/topicseen.html



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Okay guys look at this thread, watch the video and read all of it and then explain why you think it can't work.

https://overunitymachines.com/index.php/topic,61.0/topicseen.html
Thanks for the share Carroll nice work from Jim.
   
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Tinman
Quote
I took the time to find the graphs from the tests carried out in the 1+1=3 topic.
As you can see, the torque over distance increase in the modified motor, for the same input power, is greatly different.
Rotational force over distance is work being done.

I can see where this could confuse others.
It follows the notion that we should see things for what they are not what others say something should be.

For example, ChatGPT said
Quote
The statement is almost correct but needs a slight adjustment. In rotational mechanics, the product of torque (rotational force) and the angle through which it acts is the work done, not simply the rotational force over distance.

Here we can see that chatgpt is good at parroting others but cannot think creatively or in an abstract sense. For example, any small segment of the circular path could be considered as a straight line so we can say force over distance is work being done in that context. The problem is AI and many people who think like programmed machines have problems with abstract concepts. As we can see the answer depends on what level were thinking on.

For example, many people can't understand Alternating Current for the same reasons. AC is simply DC which varies in voltage/current and periodically reverses direction. Yet many make AC out to be much more than it is giving it special attributes it does not possess. AC doesn't have to be a Sine Wave or symmetrical in nature and the term only implies that the Current periodically Alternates, nothing more. Yet when we say AC all everyone can see is a Sine wave, why is that?. It's because they don't realize there mind has been programmed through repetition similar to an AI.

In fact, most learning is through education/memorization and memorizing something is basically a form of programming. We are programming our kids to think one way then judging them based on how well they can retain there programming just like an AI. :D

AC



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Okay guys look at this thread, watch the video and read all of it and then explain why you think it can't work.

https://overunitymachines.com/index.php/topic,61.0/topicseen.html

It looks easy enough to build, has anyone built one to see if the concept actually works?.

AC


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Tinman
I can see where this could confuse others.
It follows the notion that we should see things for what they are not what others say something should be.

For example, ChatGPT said
Here we can see that chatgpt is good at parroting others but cannot think creatively or in an abstract sense. For example, any small segment of the circular path could be considered as a straight line so we can say force over distance is work being done in that context. The problem is AI and many people who think like programmed machines have problems with abstract concepts. As we can see the answer depends on what level were thinking on.

For example, many people can't understand Alternating Current for the same reasons. AC is simply DC which varies in voltage/current and periodically reverses direction. Yet many make AC out to be much more than it is giving it special attributes it does not possess. AC doesn't have to be a Sine Wave or symmetrical in nature and the term only implies that the Current periodically Alternates, nothing more. Yet when we say AC all everyone can see is a Sine wave, why is that?. It's because they don't realize there mind has been programmed through repetition similar to an AI.

In fact, most learning is through education/memorization and memorizing something is basically a form of programming. We are programming our kids to think one way then judging them based on how well they can retain there programming just like an AI. :D

AC

Agree AC

It seems a little to hard for some to understand that the graphs show the energy input to the rotating mass between the measured points.

When you want to know how much energy is being put into a rotating system, you graph it at the input part of the cycle, exactly as it was done.
This then gives you a very clear vision of how much work is being done on the rotating body between the two given points.

If no energy was being transferred from the electromagnet to the PM, then no force would be seen at the PM over that distance.

So many people confuse energy with power, as we have seen here.
We are not calculating power. We are measuring a torque curve between two points (work being done) in the cycle, so as we can see the difference in energy (work being done) being put into each rotating system for a given P/In to the electromagnet.


Brad


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Agree AC

It seems a little to hard for some to understand that the graphs show the energy input to the rotating mass between the measured points.

When you want to know how much energy is being put into a rotating system, you graph it at the input part of the cycle, exactly as it was done.
This then gives you a very clear vision of how much work is being done on the rotating body between the two given points.

If no energy was being transferred from the electromagnet to the PM, then no force would be seen at the PM over that distance.

So many people confuse energy with power, as we have seen here.
We are not calculating power. We are measuring a torque curve between two points (work being done) in the cycle, so as we can see the difference in energy (work being done) being put into each rotating system for a given P/In to the electromagnet.


Brad

Hi Brad,

Your graphs do not show torque. And even if they did, it would not represent energy, or power, and I assure you I know the difference. There is confusion surrounding this so I attempted to have chatgpt help explain the details. From my first post on the subject, his answer is pasted below. It has nothing to do confusion of energy and power, or not realizing any small segment of the circular path could be considered as a straight line.
bi

Quote
User
In rotational mechanical systems, torque and energy (or work done) have the same units, newton meters, but are not the same physical quantity, correct. Explain how to avoid confusion.

ChatGPT
That's correct. Torque and energy (or work done) both have units of newton-meters (Nm), but they represent different physical quantities in rotational mechanical systems.

Torque is a measure of the rotational force applied to an object, causing it to rotate around an axis. It is typically represented by the symbol τ (tau) and is defined as the product of the force and the perpendicular distance from the axis of rotation to the line of action of the force.

Energy, or work done, in rotational systems is the product of torque and the angle through which the object rotates. It represents the amount of energy transferred to or from the system due to the rotation.

To avoid confusion between torque and energy:

Understand their definitions and physical meanings: Torque is the rotational force, while energy (work done) is the result of torque acting over a distance (angle).

Pay attention to the context: Torque is usually discussed when talking about the rotational dynamics of an object, while energy (work done) is discussed in the context of changes in rotational kinetic energy or potential energy in rotational systems.

Use appropriate symbols and terms: Use τ for torque and W for work or energy, and ensure clarity in your communication by specifying which quantity you're referring to.

By keeping these distinctions in mind and using clear terminology, you can avoid confusion between torque and energy in rotational mechanical systems.
   

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It looks easy enough to build, has anyone built one to see if the concept actually works?.

AC

Yes, I tried many years ago, and could not get it to work.
But that doesn't mean some one else won't be able to make it work.

I used a large speaker magnet, that I cut in half, and flipped one half over.
I used a slow rotating 5 inch Dimond blade and grinding paste to cut the magnet in half.
I used transformer laminate blocks instead of bolts though.

I found that because the laminated blocks were attracted to the magnet much more than they were attracted toward the curved steel stator, it got stuck at the end of the magnet halves.

That was just my findings, but i encourage everyone to try everything for your self.
Just because one person fails, does not mean the next won't succeed.

Brad


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author=bistander link=topic=4603.msg111133#msg111133 date=1711670160

Quote
Quote
Your graphs do not show torque.

As a mechanical engineer, I can tell you that that is exactly what they show.
Torque: Torque is a measure of the force that can cause an object to rotate about an axis.
!!Don't get confused between torque applied ( a force applied on a rotating body about an axis), and ft/lb of torque, or Nm of torque.!!
To clarify, a force upon a rotating body is called a torque. A force applied to a body in a linear motion, is called a force.
The graphs clearly show a measured force over distance that causes an object (the rotor) to rotate about it's axis.



 
Quote
Quote
it would not represent energy

Work = Force times Distance = Energy
Energy is not power.
Power is how fast that energy is used
As the graphs were obtained in a static manor, we can calculate the energy (force x distance), but we cannot calculate the power, as there was no motion/time.

When you compress a spring, you use force over distance.
The compressed spring now has stored energy.
That stored energy came from an applied force over distance.
Energy was transferred from an applied force over distance (work being done= energy), into the spring.
If the compressed spring now has stored energy, and energy is not a force over distance as you say, then where did the spring get it's now stored energy from?

Now, if you wanted to graph that applied force over distance that it took to compress that spring, that now has stored that energy, how would you do it ?
You would do it exactly as I did with the graphs i shared.

 
Quote
Quote
And even if they did, it would not represent energy, or power, and I assure you I know the difference. There is confusion surrounding this so I attempted to have chatgpt help explain the details. From my first post on the subject, his answer is pasted below. It has nothing to do confusion of energy and power, or not realizing any small segment of the circular path could be considered as a straight line.
bi

All I can say, and with all due respect, is to forget about what chatGPT says, and step into the real world of mechanics and mechanical engineering.

Brad


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Quote
author=bistander link=topic=4603.msg111133#msg111133 date=1711670160

Quote
As a mechanical engineer, I can tell you that that is exactly what they show.
Torque: Torque is a measure of the force that can cause an object to rotate about an axis.
!!Don't get confused between torque applied ( a force applied on a rotating body about an axis), and ft/lb of torque, or Nm of torque.!!
To clarify, a force upon a rotating body is called a torque. A force applied to a body in a linear motion, is called a force.
The graphs clearly show a measured force over distance that causes an object (the rotor) to rotate about it's axis.



 
Quote
Work = Force times Distance = Energy
Energy is not power.
Power is how fast that energy is used
As the graphs were obtained in a static manor, we can calculate the energy (force x distance), but we cannot calculate the power, as there was no motion/time.

When you compress a spring, you use force over distance.
The compressed spring now has stored energy.
That stored energy came from an applied force over distance.
Energy was transferred from an applied force over distance (work being done= energy), into the spring.
If the compressed spring now has stored energy, and energy is not a force over distance as you say, then where did the spring get it's now stored energy from?

Now, if you wanted to graph that applied force over distance that it took to compress that spring, that now has stored that energy, how would you do it ?
You would do it exactly as I did with the graphs i shared.

 
Quote
All I can say, and with all due respect, is to forget about what chatGPT says, and step into the real world of mechanics and mechanical engineering.

Brad

Hi Brad,

Thanks for the definitions. But when going from linear force distance calculation to polar torque angular, the distance is the radius, or perpendicular distance from the force vector to the center of rotation, right? You don't use that on the graphs from what I see. You state on the nice diagram:
"GRAPHS SHOW TORQUE OVER DISTANCE MEASURED BETWEEN POINT A AND B OF THE ROTATING MASS"
A to B is not a proper distance for a torque calculation.

This is similar in nature to your oversight of cosθ, the angle of applied force to the direction of motion when figuring work done in linear motion.

Quote
As the graphs were obtained in a static manor, we can calculate the energy (force x distance), but we cannot calculate the power, as there was no motion/time.

But you take a comparison of the two cases at equal input power, 12W. Without time in the mix, one case may have twice the duration, or twice the input energy. So I think you have needlessly mixed power and energy.

You're obviously welcome to run tests any way you desire. All I attempted to do was point out confusion. So carry on and good luck.
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<snip>

When you compress a spring, you use force over distance.
The compressed spring now has stored energy.
That stored energy came from an applied force over distance.
Energy was transferred from an applied force over distance (work being done= energy), into the spring.
If the compressed spring now has stored energy, and energy is not a force over distance as you say, then where did the spring get it's now stored energy from?

Now, if you wanted to graph that applied force over distance that it took to compress that spring, that now has stored that energy, how would you do it ?
You would do it exactly as I did with the graphs i shared.

Quote
energy is not a force over distance as you say

I never said that. Show me.

   

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author=bistander link=topic=4603.msg111136#msg111136 date=1711680847
Hi Brad,



Quote
But when going from linear force distance calculation to polar torque angular, the distance is the radius

The measurements were taken at the radius.
The mm stated on the graph, are arc mm, not linear mm.

Quote
GRAPHS SHOW TORQUE OVER DISTANCE MEASURED BETWEEN POINT A AND B OF THE ROTATING MASS"

As measured in the arc, not in a straight line between points A and B
The fixed line from the rotor to the scales will follow the magnets arc, and so each torque is measured through that arc, not in a straight line.

Quote
A to B is not a proper distance for a torque calculation.

It is correct when static torques are being measured, as they are always measured at the arc moment.

What is torque:  Torque is a measure of the force that can cause an object to rotate about an axis. Just as force is what causes an object to accelerate in linear kinematics, torque is what causes an object to acquire angular acceleration.

What is static torque: A static torque is one which does not produce an angular acceleration. Someone pushing on a closed door is applying a static torque to the door because the door is not rotating about its hinges, despite the force applied. Someone pedaling a bicycle at constant speed is also applying a static torque because they are not accelerating.

Quote
This is similar in nature to your oversight of cosθ, the angle of applied force to the direction of motion when figuring work done in linear motion.

There is no oversight,as the PM that the magnetic field is pulling on, will always be in the arc position. It does not, and cannot travel in, or be in a linear position.
The graphs show arc force (torque) values at each point in the arc moment, not linear values.

Quote
But you take a comparison of the two cases at equal input power, 12W. Without time in the mix, one case may have twice the duration, or twice the input energy. So I think you have needlessly mixed power and energy.

The P/in's time is a constant in both cases, as is the duration.
The duration is the distance of the graphed torque curve.
I have not mixed power and energy.
I am the one saying energy is not power.

Quote
You're obviously welcome to run tests any way you desire. All I attempted to do was point out confusion. So carry on and good luck.

I carry out static torque measurements the correct way, as has been done by engineers for many years.
Engineers carried out these very same static torque tests on steam engines, so as they could be sure that the pushrod system, crankshaft, and the wheels them self could handle the applied torque at any point in the full cycle.


Brad


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If energy is not a force over distance as you say

Quote
I never said that. Show me.

Quote reply 74
Quote
Even if it were "torque (force) over distance", it is not indicating energy or power.

Brad


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Quote reply 74
Brad

Brad,
Your reply was inside a quote so it didn't show properly here. I copied it and paste it here:
______
Quote
If energy is not a force over distance as you say

Quote
I never said that. Show me.

Quote reply 74
Quote
Even if it were "torque (force) over distance", it is not indicating energy or power.

Brad
__________

My actual statement, copied from post 74 is:
"Even if it were "torque over distance", it is not indicating energy or power."

I did not say what you claim. I wrote "torque". Not force. Not torque (force). Don't lie about what I say. And torque is not force. And I meant torque, not force.
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The measurements were taken at the radius.
The mm stated on the graph, are arc mm, not linear mm.

As measured in the arc, not in a straight line between points A and B
The fixed line from the rotor to the scales will follow the magnets arc, and so each torque is measured through that arc, not in a straight line.

It is correct when static torques are being measured, as they are always measured at the arc moment.

What is torque:  Torque is a measure of the force that can cause an object to rotate about an axis. Just as force is what causes an object to accelerate in linear kinematics, torque is what causes an object to acquire angular acceleration.

What is static torque: A static torque is one which does not produce an angular acceleration. Someone pushing on a closed door is applying a static torque to the door because the door is not rotating about its hinges, despite the force applied. Someone pedaling a bicycle at constant speed is also applying a static torque because they are not accelerating.

There is no oversight,as the PM that the magnetic field is pulling on, will always be in the arc position. It does not, and cannot travel in, or be in a linear position.
The graphs show arc force (torque) values at each point in the arc moment, not linear values.

The P/in's time is a constant in both cases, as is the duration.
The duration is the distance of the graphed torque curve.
I have not mixed power and energy.
I am the one saying energy is not power.

I carry out static torque measurements the correct way, as has been done by engineers for many years.
Engineers carried out these very same static torque tests on steam engines, so as they could be sure that the pushrod system, crankshaft, and the wheels them self could handle the applied torque at any point in the full cycle.


Brad

Where is the distance from the center of rotation used to derive the torque value from a force measurement?

I don't see that and without that I fail to comprehend how you can call a force measurement torque. 

Your helpful diagrams stated distance from A to B indicating linear. How was I to know you had used a string to get an arc around the circumference, until you just told me. From that arc distance one can get the angle, and using the angle makes all the difference. If that is the case, then the x-axis is really displacement in radians, right? And if the radius was used to calculate torque from the force measurements, I could then see a "torque curve". And as chatgpt said, "Energy, or work done, in rotational systems is the product of torque and the angle through which the object rotates." See reply 70.
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Brad,
Your reply was inside a quote so it didn't show properly here. I copied it and paste it here:
______
Quote
If energy is not a force over distance as you say

Quote
I never said that. Show me.

Quote reply 74
Quote
Even if it were "torque (force) over distance", it is not indicating energy or power.

Brad
__________

My actual statement, copied from post 74 is:
"Even if it were "torque over distance", it is not indicating energy or power."

I did not say what you claim. I wrote "torque". Not force. Not torque (force). Don't lie about what I say. And torque is not force. And I meant torque, not force.
bi

I put force in, as force is torque on a rotating body.
You continue to make the same mistake over and over, regardless of me telling you that they are one in the same when imparted on a rotating body.
I did not lie. How can you lie by placing a word that has the very same meaning as the other.

So once again--when a force is placed on a body that has linear motion, it is call a force.
When a force is placed on a rotating body, it is called a torque.

Brad


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author=bistander link=topic=4603.msg111141#msg111141 date=1711694698


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Where is the distance from the center of rotation used to derive the torque value from a force measurement?

You do not need the distance from the center of rotation to obtain the torque value being imposed on the outer rim of the rotor.
We are not measuring ft/lb or newton meters of torque. We are measuring the torque being applied to the outer rim of the rotor.
The rotor could be any diameter, but the torque imposed on the outer rim would be the same regardless.
The only thing that would change, is the ft/lb or newton meters of torque value. The torque value on the outer rim of the rotor would remain the same.
You are mixing up torque and foot pounds of torque etc. They are not the same thing.

I decided to ask your belove chatGPT what an applied force to a rotating body is called
Guess what it said
Quote
The force applied to a rotating body is called torque. Torque is a measure of the rotational force applied to an object, causing it to rotate around an axis.

I then asked it this--is an applied torque over distance energy ?
It's answer was
Quote
when torque is applied over a distance and causes an object to rotate, work is done, and energy is transferred.

I am not sure why you cannot see that we were graphing the value of torque being applied to the outer rim over a distance of a rotating body.

I said-quote: "My graphs show torque over a distance in a rotating system."

You said in post 72 -quote: That would be your graphs show torque vs angle (or radians). And that is not energy.
I once again aske your beloved chatGPT- is torque applied to a rotating disc over distance energy
It's answer-
Quote
Yes, when torque is applied to a rotating disc over a distance, it involves the transfer or conversion of energy.
 Work: When torque is applied over a distance (angular displacement), work is done on the rotating disc. Work, in this context, is the transfer of energy that results in the disc's rotational motion. The amount of work done depends on the magnitude of the torque applied and the distance over which it is applied.


So, when torque is applied to a rotating disc over a distance, it involves the transfer of energy, as work is done on the disc to make it rotate. This demonstrates the connection between torque, angular displacement, work, and energy in the context of rotating systems.

In post 74, you said:
Quote
Only the force in the direction of motion or displacement enters into the work formula.
I must say, if the torque is being imparted on the PM that is fixed to the rotating disk, in which direction do you think it was going when each measurement was taken in the graphs?
Obviously, the electromagnet, which is fixed into position, is pulling the PM on the rotor toward it, in a radial arc from point A to point B-the direction of rotation, as clearly shown in the diagram posted.

You also said in post 74 Quote:
Quote
Force over distance in rectangular coordinates or torque over angle in polar. But not a mix of the two.

I never mixed the two. It was you that was unsure how the measurements were taken, not me.

Quote
I don't see that and without that I fail to comprehend how you can call a force measurement torque.

Once again, because a force applied to a rotating body is called a torque. not foot pounds of torque, not newton meters of torque, just a torque.
That torque can be measured in grams, pounds, newtons-what ever you like,
When we were all testing this motor, we wanted to know the difference in applied torque onto the rotors outer rim during the brief on time of the coil, at the same P/in to the coil, over the same distance of rotation, between the two types of motors. By graphing the torque curve as we did, we could calculate the average torque applied to the rotor in both the motors, regardless of what diameter each persons rotor was.

Quote
Your helpful diagrams stated distance from A to B indicating linear.

No it did not.
How on earth do you get linear from a diagram that shows a magnet traveling from point A to point B on a rotor ?
The diagram says- graphs show torque over distance measured between point A and B of the !rotating mass!, which obviously means that the distance between point A and point B is the curved path of the rotating PM, which is fixed to the rotating mass.

Quote
How was I to know you had used a string to get an arc around the circumference, until you just told me.

Well how was I to know that you didn't think I new what I was doing, when i've been doing this type of thing for 40 years.
How else would you graph a torque curve using multiple static measurements over a curved path?
It would seem like you decided I was wrong, before you even knew how the test was carried out.

Quote
And if the radius was used to calculate torque from the force measurements, I could then see a "torque curve".

Once again, the radius is not needed to obtain the torque curve being imparted on the outer rim of the rotor.
The rotor could have a radius of 100mm or 500mm, but the torque being applied on the outer rim of the rotor would be exactly the same in both cases.

Once again, we refer to your beloved chatGPT
Question-what is static torque
Answer-
Static torque refers to the torque applied to an object that is not in motion, i.e., it is static or stationary. It's the torque exerted on an object without any movement or rotation occurring.
In practical terms, static torque may be important for various reasons:
Assembly and Fastening: When assembling components, especially in engineering and manufacturing, static torque is often applied to fasteners (such as nuts and bolts) to ensure they are tightened to a specific torque specification.
Calibration: Static torque is used in the calibration of torque-measuring instruments or devices, ensuring their accuracy by applying known torques to them while they are stationary.
Preload: In mechanical systems, such as bearings, static torque might be applied to create preload, which helps to reduce backlash, improve rigidity, and enhance the overall performance of the system.In summary, static torque is the torque applied to an object at rest, and it finds applications in various fields such as engineering, manufacturing, and mechanical systems.


Quote
And as chatgpt said, "Energy, or work done, in rotational systems is the product of torque and the angle through which the object rotates

Which is exactly what the graphs and diagram show.
It shows the torque value being imparted on the PM, that is fixed to the rotating disc, and so the PM will change angle slightly to that of the electromagnet, through each torque measurement point in the graph.

I really am confused as to why you find this hard to understand.
I still think you are mixing up torque with the likes of foot pounds of torque, or newton meters of torque, but they are not the same.

One more question for the almighty chatGPT

if you are pushing down on the peddle  of a pushbike, but the peddle does not move, are you applying a torque upon the peddle ?

Answer:
Quote
Yes, if you are pushing down on the pedal of a pushbike but the pedal does not move, you are indeed applying torque to the pedal. Torque is the rotational force applied to an object around an axis, and when you push down on the pedal, you are applying a force that tends to rotate the pedal around its axis of rotation.


Brad


---------------------------
Never let your schooling get in the way of your education.
   
Full Member
***

Posts: 116


You do not need the distance from the center of rotation to obtain the torque value being imposed on the outer rim of the rotor.
We are not measuring ft/lb or newton meters of torque. We are measuring the torque being applied to the outer rim of the rotor.
The rotor could be any diameter, but the torque imposed on the outer rim would be the same regardless.
The only thing that would change, is the ft/lb or newton meters of torque value. The torque value on the outer rim of the rotor would remain the same.
You are mixing up torque and foot pounds of torque etc. They are not the same thing.

I decided to ask your belove chatGPT what an applied force to a rotating body is called
Guess what it said
I then asked it this--is an applied torque over distance energy ?
It's answer was
I am not sure why you cannot see that we were graphing the value of torque being applied to the outer rim over a distance of a rotating body.

I said-quote: "My graphs show torque over a distance in a rotating system."

You said in post 72 -quote: That would be your graphs show torque vs angle (or radians). And that is not energy.
I once again aske your beloved chatGPT- is torque applied to a rotating disc over distance energy
It's answer-
In post 74, you said: I must say, if the torque is being imparted on the PM that is fixed to the rotating disk, in which direction do you think it was going when each measurement was taken in the graphs?
Obviously, the electromagnet, which is fixed into position, is pulling the PM on the rotor toward it, in a radial arc from point A to point B-the direction of rotation, as clearly shown in the diagram posted.

You also said in post 74 Quote:
I never mixed the two. It was you that was unsure how the measurements were taken, not me.

Once again, because a force applied to a rotating body is called a torque. not foot pounds of torque, not newton meters of torque, just a torque.
That torque can be measured in grams, pounds, newtons-what ever you like,
When we were all testing this motor, we wanted to know the difference in applied torque onto the rotors outer rim during the brief on time of the coil, at the same P/in to the coil, over the same distance of rotation, between the two types of motors. By graphing the torque curve as we did, we could calculate the average torque applied to the rotor in both the motors, regardless of what diameter each persons rotor was.

No it did not.
How on earth do you get linear from a diagram that shows a magnet traveling from point A to point B on a rotor ?
The diagram says- graphs show torque over distance measured between point A and B of the !rotating mass!, which obviously means that the distance between point A and point B is the curved path of the rotating PM, which is fixed to the rotating mass.

Well how was I to know that you didn't think I new what I was doing, when i've been doing this type of thing for 40 years.
How else would you graph a torque curve using multiple static measurements over a curved path?
It would seem like you decided I was wrong, before you even knew how the test was carried out.

Once again, the radius is not needed to obtain the torque curve being imparted on the outer rim of the rotor.
The rotor could have a radius of 100mm or 500mm, but the torque being applied on the outer rim of the rotor would be exactly the same in both cases.

Once again, we refer to your beloved chatGPT
Question-what is static torque
Answer-
Static torque refers to the torque applied to an object that is not in motion, i.e., it is static or stationary. It's the torque exerted on an object without any movement or rotation occurring.
In practical terms, static torque may be important for various reasons:
Assembly and Fastening: When assembling components, especially in engineering and manufacturing, static torque is often applied to fasteners (such as nuts and bolts) to ensure they are tightened to a specific torque specification.
Calibration: Static torque is used in the calibration of torque-measuring instruments or devices, ensuring their accuracy by applying known torques to them while they are stationary.
Preload: In mechanical systems, such as bearings, static torque might be applied to create preload, which helps to reduce backlash, improve rigidity, and enhance the overall performance of the system.In summary, static torque is the torque applied to an object at rest, and it finds applications in various fields such as engineering, manufacturing, and mechanical systems.


Which is exactly what the graphs and diagram show.
It shows the torque value being imparted on the PM, that is fixed to the rotating disc, and so the PM will change angle slightly to that of the electromagnet, through each torque measurement point in the graph.

I really am confused as to why you find this hard to understand.
I still think you are mixing up torque with the likes of foot pounds of torque, or newton meters of torque, but they are not the same.

One more question for the almighty chatGPT

if you are pushing down on the peddle  of a pushbike, but the peddle does not move, are you applying a torque upon the peddle ?

Answer:

Brad

Brad,
Sincerely, thank you the discussion. But I will need to leave it. A basic disagreement exists I fail to see resolution. Your last question:
Quote
if you are pushing down on the peddle  of a pushbike, but the peddle does not move, are you applying a torque upon the peddle

My answer is: No. You are applying force to the peddle. Torque occurs on the crank shaft through the moment arm.
Torque is not force.

Regards.
bi

   
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