[hamdani yusuf (OP)]

No mention of angle, anywhere.

The angle is implied by the usage of rotational radius.

But neither dictionary definition mentions "rotational radius".

If I hang a 1N weight on a 1 m arm and it moves 1 microradian, what is the torque?

If torque is a cause, expressed in newton meters, τ = 1Nm, as measured by any torque meter

If torque is an effect, expressed in newton meters per radian, τ = 1,000,000 Nm/rad according to Hamdani.

Which is the more useful number?

Suppose it moves 2 μrad

Then conventionally τ = 1 Nm

But using your definition, τ = 500,000 Nm/rad

So why does the torque wrench read the same value?

In my unit, the meter represents arc length of the rotational displacement. Your calculation shows that you are confusing it with the radius of rotation.

[hamdani yusuf (OP)]

My proposal to solve the inconsistency problem in current standard units of rotational quantities is to distinguish between geometric and rotational radius. In this Wikipedia article, it's called Radius of gyration.
https://en.wikipedia.org/wiki/Radius#See_also

The radius of gyration or gyradius of a body about the axis of rotation is defined as the radial distance to a point which would have a moment of inertia the same as the body's actual distribution of mass, if the total mass of the body were concentrated there. The radius of gyration has dimensions of distance [L] or [M0LT0] and the SI unit is the metre (m).

Mathematically the radius of gyration is the root mean square distance of the object's parts from either its center of mass or a given axis, depending on the relevant application. It is actually the perpendicular distance from point mass to the axis of rotation. One can represent a trajectory of a moving point as a body. Then radius of gyration can be used to characterize the typical distance travelled by this point.
https://en.wikipedia.org/wiki/Radius_of_gyration

In case you already forget, below is my previous post about it. Before we move further, I want to know if there any objection to the equations I put in the table? Can we agree on their validity?

I'm quite happy dealing with people who don't know much physics, but not with someone who refuses to learn.

Between these two tables, which one is more consistent?

You don't seem to be aware of the inconsistency in current standard units of rotational quantities, as shown clearly in this table.


Compare them with the new proposed standard units, which are consistent with the relating equations.

If you still wonder why some people have proposed changes to current standard units in rotational quantities, read the tables above thoroughly, and understand what they mean and imply.

Before we debate about the difference, let's discuss about their similarities first.
The proposed new standard units are the same as current standard units for angle, angular velocity, and angular acceleration. Also for work and power. Units for kinetic energy and potential energy are also still the same.

Now let's take a look at their difference. First, rotational inertia, I.
The current standard unit for I is kg.m^2, which is based on the equation I = mass times rotational radius squared.
I = m R^2
The current standard unit for mass is kilogram, while the current standard unit for rotational radius is meter.
But don't forget that rotational inertia is also involved in many other equations.
I = L/ω
I = τ/α
I = 2 Ek / ω^2
Let's use the last equation, because we have the same agreed units for kinetic energy and angular velocity, thus there should be no dispute to the result.
Standard unit for kinetic energy is kg m^2 s^-2
Standard unit for angular velocity is rad s^-1,
thus ω^2 has standard unit rad^2 s^-2
Thus the Standard unit for rotational inertia should be kg m^2 rad^-2

Now we need to reconcile with the equation used for the current standard unit for rotational inertia.
I = m R^2 = 2 Ek / ω^2
R^2 = I/m
R = √(I/m)
Thus the standard unit for rotational radius should be meter per radian.

[paul cotter]

NONSENSE, as has been explained to you countless times.

[alancalverd]

In my unit, the meter represents arc length of the rotational displacement.

So what do you call the quantity that a torque meter measures, or a torque wrench delivers?

If I want to use the same force to apply more torque, do I use a longer lever, or keep the lever the same length and move it further?

You seem confused by the fact that torque is dimensionally identical to energy but is intentionally expressed in Nm, not J, to distinguish it. Most people find it amusing, not confusing.

Standard unit for kinetic energy is kg m^2 s^-2
Standard unit for angular velocity is rad s^-1,

No. Dimensions are not units. The unit of energy is the joule.

[Bored chemist]

And re this "
Let's compare facts and figures. How many control loops have you designed, implemented, commissioned, improved, tuned in the last twenty years? "
For the sake of discussion (And to save the trouble  of estimating a count) , let's pretend it is just one. Let's say it's a system just like the one in the video about a sensitive balance

The point is that I understood that system and it seems you do not.

You misunderstood that system. That's why you think someone else who disagree with you doesn't.

Nice claim.
Post evidence.

Re "Here's another example to show that expected rotational radius "
That depends whether or not you are sensible in your expectations.
If you are not, that isn't our responsibility.
This "The question is, what's the torque produced by the force?" is a meaningless question.
It's like asking "How far is it to Rome?". There is no "right" answer.

If you don't specify "about such and such a point" then you can't sensibly ask what the torque is.

Pissing about with the units does not alter that.



And, if I have measured the length of the green line in this picture correctly, the answer is about 0.55 N m

Diag 2.png (15.27 kB . 722x588 - viewed 485 times)

If you think that diagram is a video about a sensitive balance then you really have lost the plot.

Your diagram clearly shows your misunderstanding of the concept.

My diagram shows that you failed to specify things properly.
"This "The question is, what's the torque produced by the force?" is a meaningless question.
It's like asking "How far is it to Rome?". There is no "right" answer.

If you don't specify "about such and such a point" then you can't sensibly ask what the torque is."

So, once again, that's a failure on your part, not mine.

[Bored chemist]

Not many members are active in this forum.

Guess why.

[hamdani yusuf (OP)]

NONSENSE, as has been explained to you countless times.

Can you point out the best explanation that you know?
Do you agree that all of the equations in the tables are correct?

[hamdani yusuf (OP)]

In my unit, the meter represents arc length of the rotational displacement.

So what do you call the quantity that a torque meter measures, or a torque wrench delivers?

If I want to use the same force to apply more torque, do I use a longer lever, or keep the lever the same length and move it further?

You seem confused by the fact that torque is dimensionally identical to energy but is intentionally expressed in Nm, not J, to distinguish it. Most people find it amusing, not confusing.

Torque.

use a longer lever. It will make the displacement arclength proportionally longer with the same displacement angle as you apply the same force.

Do you think Nm has the same dimension as Nm/rad?

[hamdani yusuf (OP)]

Standard unit for kinetic energy is kg m^2 s^-2
Standard unit for angular velocity is rad s^-1,

No. Dimensions are not units. The unit of energy is the joule.

Don't you think that
1 Joule = 1 kg m^2 s^-2 ?
5 Joule = 5 kg m^2 s^-2 ?
99 Joule = 99 kg m^2 s^-2 ?

[hamdani yusuf (OP)]

And re this "
Let's compare facts and figures. How many control loops have you designed, implemented, commissioned, improved, tuned in the last twenty years? "
For the sake of discussion (And to save the trouble  of estimating a count) , let's pretend it is just one. Let's say it's a system just like the one in the video about a sensitive balance

The point is that I understood that system and it seems you do not.

You misunderstood that system. That's why you think someone else who disagree with you doesn't.

Nice claim.
Post evidence.

Re "Here's another example to show that expected rotational radius "
That depends whether or not you are sensible in your expectations.
If you are not, that isn't our responsibility.
This "The question is, what's the torque produced by the force?" is a meaningless question.
It's like asking "How far is it to Rome?". There is no "right" answer.

If you don't specify "about such and such a point" then you can't sensibly ask what the torque is.

Pissing about with the units does not alter that.



And, if I have measured the length of the green line in this picture correctly, the answer is about 0.55 N m

Diag 2.png (15.27 kB . 722x588 - viewed 485 times)

If you think that diagram is a video about a sensitive balance then you really have lost the plot.

Your diagram clearly shows your misunderstanding of the concept.

My diagram shows that you failed to specify things properly.
"This "The question is, what's the torque produced by the force?" is a meaningless question.
It's like asking "How far is it to Rome?". There is no "right" answer.

If you don't specify "about such and such a point" then you can't sensibly ask what the torque is."

So, once again, that's a failure on your part, not mine.

My original diagram is a response to claim that torque can be easily determined by equation force times rotational radius. It shows that even when the force can be accurately measured, you still need to measure the actual rotational radius, which is the distance between the axis of rotation and the point where the force is applied to the system. I showed that in some cases the location of rotational axis can't be reliably determined before actual rotational displacement occurs, no matter how small it is, as long as it's non-zero.
Your arguments above basically agree with mine, although you don't seem to realize it.

[hamdani yusuf (OP)]

Not many members are active in this forum.

Guess why.

Perhaps they had some unpleasant experiences when expressing their opinions here from the replies that they got from some other members. Or they are afraid to get the same feeling from similar experience that those members had.

[hamdani yusuf (OP)]

If the bolt freely turns for 1 microradian then the spanner will not hit the obstruction and the torque will be 1Nm. The bolt turns 1 microradian, sin1microradian = 1x10 exp -6,   0.5x 1x10 exp -6 =0.5x10 exp -6,  this is half the distance to the obstructing block.  Edit: bullshit, you do not need tensor analysis for simple mechanics.

So, is it a cancellation or counterbalance?

As shown in the video, if you can solve the problem without using tensor analysis, you can consider yourself lucky, because most of the parameters are zero.

It's possible for you to distinguish between cancellation and counterbalance by the reading of the measuring device. If the measurement shows 0, you call that cancellation. If it shows the same value as the calculation, you call it counterbalance. In between, you can call it partial cancellation.

[hamdani yusuf (OP)]

Torque wrench test! (Proof that hand position REALLY matters.) | Auto Expert John Cadogan

This video explores the importance of hand position when using a torque wrench. The presenter demonstrates how hand position affects the torque applied to a fastener, using a torque transducer to measure the actual torque applied. The presenter emphasizes that using a torque wrench incorrectly can lead to over-tightening or under-tightening fasteners.

[hamdani yusuf (OP)]

I'm preparing my next video on standard units of rotational quantities. It will delve deeper into the proposal to distinguish between geometric and rotational radius. In many cases, the axis of rotation is not easily determined or observed. It needs to be calculated from the arc length of two adjacent points on the curve and their difference in angle of normal lines.
https://en.wikipedia.org/wiki/Radius_of_curvature


Some other examples in orbital mechanics.

[paul cotter]

The behaviour of orbital bodies has nothing to do with torque.

[hamdani yusuf (OP)]

The behaviour of orbital bodies has nothing to do with torque.

That's because you have limited your own definition of torque. But the mathematical equations are definitely there showing that they are related. Whenever a system change its angular momentum, there must be some torque involved. Control of satellites and spacecrafts are practical problems in this field. They are examples where rotational radius isn't necessarily constant.
Pendulums are some other examples.

Conical pendulum


Foucault pendulum

[Bored chemist]

I'm not quoting the whole of reply 949 , it's already too long.
But here's part of it "I showed that in some cases the location of rotational axis can't be reliably determined before actual rotational displacement occurs".

Then how does the thing, in real life, "know" how to move?

Are you saying that the use of a torque wrench is non deterministic?

The fact is that a system with forces acting on it has many different torques depending on what point you consider to be the axis.
That's the point behind my observation about it depending if you are sitting on the bench.

Now, in other case where there is no actual rotation, even so slightly, what is the radius of rotation that you will use to calculate torque?

It depends if I'm sat on the bench or stood next to it.

You did not understand it then.

[hamdani yusuf (OP)]

I'm not quoting the whole of reply 949 , it's already too long.
But here's part of it "I showed that in some cases the location of rotational axis can't be reliably determined before actual rotational displacement occurs".

Then how does the thing, in real life, "know" how to move?

Are you saying that the use of a torque wrench is non deterministic?

The fact is that a system with forces acting on it has many different torques depending on what point you consider to be the axis.
That's the point behind my observation about it depending if you are sitting on the bench.

Now, in other case where there is no actual rotation, even so slightly, what is the radius of rotation that you will use to calculate torque?

It depends if I'm sat on the bench or stood next to it.

You did not understand it then.

You basically ask to determine whether the jammed thread would break before measuring it first.
Things in real life simply react to changes in their environment. You just can't predict their reaction if you don't know the conditions significantly affect their environment through chain of causes and effects. The system is basically deterministic. It's just underdetermined when some important parameters are still missing, not measured or calculated yet.
If you are sitting on the bench, you apply force to the bench in the form of body weight and reactionary force from the wrench that you push upward. If you are standing on the floor, those forces will be applied to the floor instead of the bench.

[Bored chemist]

You basically ask to determine whether the jammed thread would break before measuring it first.

I did not ask anything of the sort. That's a misunderstanding on your part.

I pointed out that, if I was stood on the floor, the axis of rotation would be different from if I was sat on the bench.

And you still don't understand, but you still assume I'm wrong.

[hamdani yusuf (OP)]

You basically ask to determine whether the jammed thread would break before measuring it first.

I did not ask anything of the sort. That's a misunderstanding on your part.

I pointed out that, if I was stood on the floor, the axis of rotation would be different from if I was sat on the bench.

And you still don't understand, but you still assume I'm wrong.

Not necessarily. If the thread is loose enough, or the bench is heavy enough, your position won't change the axis of rotation.