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Topic: Torque , general question (Read 467 times) previous topic - next topic

chris_abc

#15
Feb 22, 2021, 05:30 pm Last Edit: Feb 22, 2021, 05:31 pm by chris_abc
Just a quick point I wanted to add so that you don't make a mistake that I see people often make in this type of problem (and have made my self as a mech. engineering student):

The distance between your gravity vector to your motor axis is the actual arm you are looking for. Notice that this distance changes depending on where the person is w.r.t. the motor.

So the distance you are looking for is as stated in the picture: A=sin(θ)*r
NOT A = r

Imagine rotating your system by hand instead of with the motor, I think you can imagine that it gets heavier to do the bigger theta becomes, this is due to your arm becoming larger and larger for a larger theta (up until theta is 90 degrees, after that it becomes smaller again). In fact, if theta is 90 degrees, sin(θ)=1 and the formula does become M=r*F, but ONLY for theta=90 degrees. Let me know if this confuses you because it's quite important to grasp this before you go further I think.

Another note of safety: Make sure the seat of the person is located so that the COM of the person is BELOW the motor axis, such that if there is a motor failure, the person will simply swing back to the position theta = 0. If the COM were above the axis, the person would fall over in case of a motor failure.
 

Robin2

Just a quick point I wanted to add so that you don't make a mistake that I see people often make in this type of problem (and have made my self as a mech. engineering student):
I can't relate your diagram to the OP's diagram in Reply #12

...R
Two or three hours spent thinking and reading documentation solves most programming problems.

quebeclima

#17
Feb 22, 2021, 06:25 pm Last Edit: Feb 22, 2021, 06:26 pm by quebeclima
Just a quick point I wanted to add so that you don't make a mistake that I see people often make in this type of problem (and have made my self as a mech. engineering student):

The distance between your gravity vector to your motor axis is the actual arm you are looking for. Notice that this distance changes depending on where the person is w.r.t. the motor.

So the distance you are looking for is as stated in the picture: A=sin(θ)*r
NOT A = r

Imagine rotating your system by hand instead of with the motor, I think you can imagine that it gets heavier to do the bigger theta becomes, this is due to your arm becoming larger and larger for a larger theta (up until theta is 90 degrees, after that it becomes smaller again). In fact, if theta is 90 degrees, sin(θ)=1 and the formula does become M=r*F, but ONLY for theta=90 degrees. Let me know if this confuses you because it's quite important to grasp this before you go further I think.

Another note of safety: Make sure the seat of the person is located so that the COM of the person is BELOW the motor axis, such that if there is a motor failure, the person will simply swing back to the position theta = 0. If the COM were above the axis, the person would fall over in case of a motor failure.
 
I think I understand what you're saying, but I believe your example only applies when the COM is not directly in the axis of the motor right?
Ideally I would try to balance out the structure so the COM coincides with the motor axis, basically eliminating the influence gravity has.

chris_abc

I think I understand what you're saying, but I believe your example only applies when the COM is not directly in the axis of the motor right?
Ideally I would try to balance out the structure so the COM coincides with the motor axis, basically eliminating the influence gravity has.
In theory, if your COM is exactly in line with the motor axis, you are correct in that the influence of gravity will be zero. However I doubt you can get it to be exactly in line, as modelling a person is quite hard (everyone is different), and the thing you are modelling a COM of will also move and alter it's own geometry (i.e. the person will move their head, their arms, or even their legs), which would change your COM.

If I had to choose between the option of designing my COM slightly under the axis and needing a slightly heavier motor for that, or the risk of the person in the device raising their arm and thereby toppling over, I would definitely go for the first one.
Especially considering that you will need quite a strong motor anyway, if you want to move around rotate a person at a reasonable speed. Coming back to moment of inertia: a heavier, larger object will be harder to start rotating then, for example, a lightweight plastic globe.

quebeclima

#19
Feb 23, 2021, 10:20 am Last Edit: Feb 23, 2021, 10:21 am by quebeclima
In theory, if your COM is exactly in line with the motor axis, you are correct in that the influence of gravity will be zero. However I doubt you can get it to be exactly in line, as modelling a person is quite hard (everyone is different), and the thing you are modelling a COM of will also move and alter it's own geometry (i.e. the person will move their head, their arms, or even their legs), which would change your COM.

If I had to choose between the option of designing my COM slightly under the axis and needing a slightly heavier motor for that, or the risk of the person in the device raising their arm and thereby toppling over, I would definitely go for the first one.
Especially considering that you will need quite a strong motor anyway, if you want to move around rotate a person at a reasonable speed. Coming back to moment of inertia: a heavier, larger object will be harder to start rotating then, for example, a lightweight plastic globe.
Ok then we're on the same page. I agree with the fail-safe of having the CG slightly below the axis. I'm looking at a motor of 8 to 12NM, maybe have to add a gearbox, but I'm trying to gather as much info as possible to make sure calculation wise (as far as I can) it's correct.

quebeclima

#20
Feb 23, 2021, 03:16 pm Last Edit: Feb 23, 2021, 04:48 pm by quebeclima Reason: found some formula's which might be a solution to my issue
Just a quick point I wanted to add so that you don't make a mistake that I see people often make in this type of problem (and have made my self as a mech. engineering student):

The distance between your gravity vector to your motor axis is the actual arm you are looking for. Notice that this distance changes depending on where the person is w.r.t. the motor.

So the distance you are looking for is as stated in the picture: A=sin(θ)*r
NOT A = r

Imagine rotating your system by hand instead of with the motor, I think you can imagine that it gets heavier to do the bigger theta becomes, this is due to your arm becoming larger and larger for a larger theta (up until theta is 90 degrees, after that it becomes smaller again). In fact, if theta is 90 degrees, sin(θ)=1 and the formula does become M=r*F, but ONLY for theta=90 degrees. Let me know if this confuses you because it's quite important to grasp this before you go further I think.

Another note of safety: Make sure the seat of the person is located so that the COM of the person is BELOW the motor axis, such that if there is a motor failure, the person will simply swing back to the position theta = 0. If the COM were above the axis, the person would fall over in case of a motor failure.
 
if you don't mind, just to come back to this,. You mention that the moment is A*F = sin(θ)*r*F.
To see if I get it right:
I understand we have to take the sin() of the angle otherwise we're ignoring basic trigonometry, and as a result we'll obtain the arm to use in the calculation for torque/inertia?


Now when looking at the moment of inertia examples on following site, they're using 2D shapes to calculate the moment of inertia and another formula in case the centroids don't align.





I'm wondering if you could explain me what to do in case off a 3D shape.  Am I correct in thinking that I have to look at the cross-section on the x-axis, calculate the inertia of each shape and then look at the cross-section off the y-axis and calculate the inertia off each shape. Do I just add them up?


EDIT: I've done some reading and as I understand it there's two different types of inertia I was looking at:
- Area moment of inertia;
- Mass moment of inertia;


The latter being the one that applies to my project to get an idea of torque.

As I see it I just have to create worst case scenario where the body is in a position that creates the biggest inertia and calculate from there.

The way I'm going to do this is by creating a 'human' out off shapes which have existing formula's to calculate inertia and then I believe I'm set regarding the minimum torque to which I'd have to apply a safety factor.


chris_abc

#21
Feb 23, 2021, 04:53 pm Last Edit: Feb 23, 2021, 04:55 pm by chris_abc
if you don't mind, just to come back to this,. You mention that the moment is A*F = sin(θ)*r*F.
 To see if I get it right:
I understand we have to take the sin() of the angle otherwise we're ignoring basic trigonometry, and as a result we'll obtain the arm to use in the calculation for torque/inertia?
Now when looking at the moment of inertia examples on following site, they're using 2D shapes to calculate the moment of inertia and another formula in case the centroids don't align.


I'm wondering if you could explain me what to do in case off a 3D shape.  Am I correct in thinking that I have to look at the cross-section on the x-axis, calculate the inertia of each shape and then look at the cross-section off the y-axis and calculate the inertia off each shape.


Do I then just make the sum off inertia's?

Calculating a moment of inertia of a shape as complicated as that of a human is quite involved. As you can see on the website you linked, the more complicated the shape becomes, the more complicated the moment of inertia is.
But luckily, you don't need to calculate the exact moment of inertia of a person, as you should take some safety margin in the choice of your motor, hence an idea of the order of magnitude of this inertia is enough for a first calculation. Then you can see if the inertia is really the limiting factor here, or if it can be neglected due to the torque needed to overcome gravity being much larger (I am not sure which is the case here, this also depends on how quickly you want to be able to move the person around, so you should get any idea of how fast that would be.)

Now to calculate the inertia, you are on the right track that you should calculate it around a certain axis. But you should only do that around one axis: your motor axis. (I am not sure on which directions you mean with x-axis and y-axis) I would simply assume that the person in the chair can be modeled as a uniform circle, weight distribution-wise (The weight distribution is basically what you are looking for when you are talking about inertia). Of course this is an assumption that is not 100% true to reality, but it's simple and good enough.
Now to calculate your moment of inertia, determine the inertia of the circle that models the person, set this as I ( I=(pi/2)*R^4 ), and add to this the area of the circle multiplied by the area of the circle multiplied by the distance between the centroid (COM) and your motor axis squared: A*r^2

Now the moment of inertia is I' = I + A*r^2 = (pi/2)*R^4 + pi*R^2 * r^2
In which r is the distance between the COM , and R is the radius of the circle that models the person.
I find that this website explains it really well, starting with simple shapes: http://hyperphysics.phy-astr.gsu.edu/hbase/tdisc.html#:~:text=The%20moment%20of%20inertia%20of,cylinder%20about%20an%20end%20diameter.

Now you have what is actually called "Area Moment of Inertia"in [m^4] , but you need it in [kg*m^2] to use it in the formula that we want for the angular acceleration: torque / Im = theta_dd (dd = double dot, for second derivative of theta -> angular acceleration), this is called the "Mass Moment of Inertia". Which you can relate pretty simply by realizing that the circle we used to model the person actually has a certain mass and a certain area, hence we can calculate the mass/area -> [kg/m^2].
Im can be calculated pretty simply from I' :
Im = I'*M/A, with unit [kg/m^2]

Hope this helps. It's important to realize that the gravity component and the inertia component are really separate: the r is used for both, but the function with sin really has nothing to do with the inertia. The inertia also won't change depending on our theta.

quebeclima

#22
Feb 23, 2021, 05:06 pm Last Edit: Feb 23, 2021, 05:08 pm by quebeclima
Calculating a moment of inertia of a shape as complicated as that of a human is quite involved. As you can see on the website you linked, the more complicated the shape becomes, the more complicated the moment of inertia is.
But luckily, you don't need to calculate the exact moment of inertia of a person, as you should take some safety margin in the choice of your motor, hence an idea of the order of magnitude of this inertia is enough for a first calculation. Then you can see if the inertia is really the limiting factor here, or if it can be neglected due to the torque needed to overcome gravity being much larger (I am not sure which is the case here, this also depends on how quickly you want to be able to move the person around, so you should get any idea of how fast that would be.)

Now to calculate the inertia, you are on the right track that you should calculate it around a certain axis. But you should only do that around one axis: your motor axis. (I am not sure on which directions you mean with x-axis and y-axis) I would simply assume that the person in the chair can be modeled as a uniform circle, weight distribution-wise (The weight distribution is basically what you are looking for when you are talking about inertia). Of course this is an assumption that is not 100% true to reality, but it's simple and good enough.
Now to calculate your moment of inertia, determine the inertia of the circle that models the person, set this as I ( I=(pi/2)*R^4 ), and add to this the area of the circle multiplied by the area of the circle multiplied by the distance between the centroid (COM) and your motor axis squared: A*r^2

Now the moment of inertia is I' = I + A*r^2 = (pi/2)*R^4 + pi*R^2 * r^2
In which r is the distance between the COM , and R is the radius of the circle that models the person.
I find that this website explains it really well, starting with simple shapes: http://hyperphysics.phy-astr.gsu.edu/hbase/tdisc.html#:~:text=The%20moment%20of%20inertia%20of,cylinder%20about%20an%20end%20diameter.

Now you have what is actually called "Area Moment of Inertia"in [m^4] , but you need it in [kg*m^2] to use it in the formula that we want for the angular acceleration: torque / Im = theta_dd (dd = double dot, for second derivative of theta -> angular acceleration), this is called the "Mass Moment of Inertia". Which you can relate pretty simply by realizing that the circle we used to model the person actually has a certain mass and a certain area, hence we can calculate the mass/area -> [kg/m^2].
Im can be calculated pretty simply from I' :
Im = I'*M/A, with unit [kg/m^2]

Hope this helps. It's important to realize that the gravity component and the inertia component are really separate: the r is used for both, but the function with sin really has nothing to do with the inertia. The inertia also won't change depending on our theta.

I feel it's all slowly coming together.
I initially didn't realise there was a difference between area moment of inertia and mass moment of inertia (updated my reply #20 when I found out but in the mean time you replied as well).
I also didn't fully understand the relation between the two but your explanation actually made that link for me.
This is definitely something I can work with for now.

Thank you for your time and effort in explaining, genuinely appreciate it!


MarkT

"Area moment of inertia" is rather a misnomer as its nothing to do with inertia at all. 
Its correctly called "2nd moment of area" in mathematics and physics, however the engineering
textbooks are much sloppier with their nomenclature it seems...

"Mass moment of inertia" is a tautology of course, its simply "moment of inertia" or MoI.
[ I DO NOT respond to personal messages, I WILL delete them unread, use the forum please ]

chris_abc

"Area moment of inertia" is rather a misnomer as its nothing to do with inertia at all.  
Its correctly called "2nd moment of area" in mathematics and physics, however the engineering
textbooks are much sloppier with their nomenclature it seems...

"Mass moment of inertia" is a tautology of course, its simply "moment of inertia" or MoI.
Try connected them by mass/area , they definitely have a strong relationship!

chris_abc

I feel it's all slowly coming together.
I initially didn't realise there was a difference between area moment of inertia and mass moment of inertia (updated my reply #20 when I found out but in the mean time you replied as well).
I also didn't fully understand the relation between the two but your explanation actually made that link for me.
This is definitely something I can work with for now.

Thank you for your time and effort in explaining, genuinely appreciate it!


You're welcome, good luck with the project.

quebeclima


If I need to create a seperate thread, please let me know.


Progress
I've done some calcuations and worked as follows:


- I created the basic 'human' shape and the rough shapes for the structure based on the shapes I could find for calculating mass moment of inertia

This leaves me with inertia's per bodypart and structureparts.

then I calculated the torque required to overcome this inertia with the formula T = I .α where: T = Torque, I = inertia and α is angular acceleration.


- I then calculated the torque which was the initial subject of this thread, namely T = F.r where T = torque, F = force and r = radius from the rotation axis.




Follow-up question  
To give you an idea off how I'm calculating, below is an image to illustrate, top left are the basic shapes mentioned earlier, top right is the structure which needs calculation. Middle is the 'person' in front view with the CG's for each bodypart.

The 'human' is created using average weights and average measurements found online. Not the pretiest, but it's good enough for my calculations. The CG is calculated using a program.

inertia

As for the questions:
A] Can I just make the sum of all the torque calculated to get the total torque required (both inertia and the other one) or is this too simplistic?

B] not focussing on the torque required to overcome inertia but rather on the other one (T = F.r). 
Assuming I can indeed make the sum of all off the calculated torque, I end up with around 158 Nm.  


In my first post my issue was understanding how torque translated in practice. 
I understand the concept that if I have a 8Nm motor, this means I can lift 8N (or .8kg) at a distance off 1 meter. 
At 0.1 m this would mean a force of 80N (or 8Kg).


With the above info, does it mean that the resulting calculated 158 Nm needs to be overcome by a theoretical motor of 158 Nm (in reality motor + gearing), or is this 158 N at the distance calculated from the axis meaning I can for example take  a 12Nm motor with a lever of let's say 1 cm to have 1200N at 1cm?


If I'm confusing you please let me know, I've been working a long time on it so it's obvious for me but I understand it might not be as clear to you the way I'm explaining it (non-native english speaker, so please bare with me) 


rgds

MarkT

#27
Mar 03, 2021, 09:54 pm Last Edit: Mar 03, 2021, 09:54 pm by MarkT
A) Torques add linearly (for the same axis)

B) yes your motor/gearing system has to be able to handle the max load torque or it will fail.
The better the balance is the less wear and tear will happen too, so its good to try to make
the commonest case much better than the worst case.

And be warned these are large torques, you probably want to limit the speeds involved
for safety, this sort of torque can break your arm for instance.

My rough guide to torque in human terms:

0.1Nm - finger and thumb twisting
1Nm - tightening a screwdriver
10Nm - steering wheel, both hands
100Nm - hazardous to life and limb if you try to tackle it.
[ I DO NOT respond to personal messages, I WILL delete them unread, use the forum please ]

quebeclima

B) yes your motor/gearing system has to be able to handle the max load torque or it will fail.
The better the balance is the less wear and tear will happen too, so its good to try to make
the commonest case much better than the worst case.
Just to confirm, the 12 Nm motor would NOT be enough to overcome the calculated torque, right?

I'd need to find a gearing system? (the part that confuses me is that I have 1200N at 1cm but the motor is rated 12Nm and I calculated 158Nm (N vs Nm) ) 
Quote
And be warned these are large torques, you probably want to limit the speeds involved
for safety, this sort of torque can break your arm for instance.

My rough guide to torque in human terms:

0.1Nm - finger and thumb twisting
1Nm - tightening a screwdriver
10Nm - steering wheel, both hands
100Nm - hazardous to life and limb if you try to tackle it.
I see, It's not easy getting a picture off the magnitude since I only know of torque by the numbers and what I read.
This certainly helps to get an idea! 
I just picked an arbitrary number for angular acceleration of 8 rad/s² for my calculations, based on simulator data, but I can always tone this down. 

MarkT

#29
Mar 04, 2021, 01:42 pm Last Edit: Mar 04, 2021, 01:43 pm by MarkT
Just to confirm, the 12 Nm motor would NOT be enough to overcome the calculated torque, right?
If you calculated 158Nm and that's actually right, no, obviously not...
Quote
I'd need to find a gearing system?
You need enough torque, one way or another.
Quote
(the part that confuses me is that I have 1200N at 1cm but the motor is rated 12Nm and I calculated 158Nm (N vs Nm) )
1200N at 1cm is 12Nm
Have no idea at all why you mean by " I calculated 158Nm (N vs Nm)" - if you calculate
torque its in Nm.

I'm now wondering how you did your calculations....

Quote
I see, It's not easy getting a picture off the magnitude since I only know of torque by the numbers and what I read.
This certainly helps to get an idea!
I just picked an arbitrary number for angular acceleration of 8 rad/s² for my calculations, based on simulator data, but I can always tone this down.
If you need significant angular accelerations like that you'll need to have an estimate of MoI
too, as acceleration torque may actually be dominant for such large accelerations.

I'm beginning to worry that you've no idea of the safety implications of such a machine, for instance
you may be into whiplash territory at 8 rad/s/s...
[ I DO NOT respond to personal messages, I WILL delete them unread, use the forum please ]

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