# Lead screw's moment of inertia

## Summary:

calculation of lead screw moment inertia with weight

## Main Question or Discussion Point

In order to choose a DC motor according to the article titled: APPYING MOTORS IN LINEAR MOTION APPLICATION by PITTMAN -
step 4 : "determine the total reflected inertia (Jt) back from the load to lead screw shaft " .
The formula is: Jt = Jscrew + Jload.

This calculation relies on the fact that the load moves on a power screw - see the attached schema - in Scheme # 2.

My system is different and the number in the schema is # 1
The load is located at distance L, on both sides of the center of the screw.

My request: How do you calculate the load inertia in this situation?

#### Attachments

• 2.1 MB Views: 31
• 118.2 KB Views: 30

Related General Engineering News on Phys.org
Welcome, beashayyael Perhaps I don't understand the problem correctly, but the following seems to be contradictory:

A) The article shows a mechanism that must deal with the inertia of the screw and the load, which is a weight that the screw forces to translate back and forth.

B) Your schematic #2 shows a shaft that must rotate a weight that is located off the center of rotation, presenting rotational inertia to the motor. In this specific case, the pitch or mechanical advantage of the lead screw is irrelevant, since that weight is solidly attached to it.

C) The two weights represented as attached to the orange shaft will rotate around that shaft in similar way? If so, the calculation should consider the MA of the screw, as well as the reduced angular velocity of those weights respect to case B above. In practical terms, the lead screw-nut assembly may encounter additional friction and resistance from any unbalanced torque coming from the unequal displacement of the bearings located at the ends of that orange shaft.

jrmichler
Mentor
The original post is unclear. Some basic principles of servo systems:

1) Everything that moves when the motor runs has an inertia that is reflected back to the motor. It helps to clearly label things that rotate and things that translate.
2) All bearings add friction that is reflected back to the motor.
3) If a mass has a vertical component of motion, the force to move against gravity is reflected back to the motor.

In order to solve this problem, show all of the above on one or more diagrams. If a mass is moving in translation, search terms ball screw load inertia will lead to the information to calculate the inertia reflected to the motor.

Each acceleration, friction, and gravity force is independent, so you calculate each individually. Then add them up in the final step.

• beashayyael and Lnewqban
The original post is unclear. Some basic principles of servo systems:

1) Everything that moves when the motor runs has an inertia that is reflected back to the motor. It helps to clearly label things that rotate and things that translate.
2) All bearings add friction that is reflected back to the motor.
3) If a mass has a vertical component of motion, the force to move against gravity is reflected back to the motor.

In order to solve this problem, show all of the above on one or more diagrams. If a mass is moving in translation, search terms ball screw load inertia will lead to the information to calculate the inertia reflected to the motor.

Each acceleration, friction, and gravity force is independent, so you calculate each individually. Then add them up in the final step.

I am attaching a new scheme that on the last page is drawn a bearing with the forces that exert on it.
The weight that the screw sees is the one that appears in the diagram as
"W * cos"?

#### Attachments

• 172.8 KB Views: 30
jrmichler
Mentor
Be careful, because weight is not the same as mass. When accelerating a mass in any direction, the acceleration force is the actual mass. The force due to gravity depends on the direction of motion, which can be horizontal to vertical, and up or down. A good reference, found using search criteria ball screw load inertia is this from THK: https://tech.thk.com/en/products/pdf/en_b15_069.pdf. If you are moving the load horizontally, it has full mass, but (as far as the ball screw is concerned) zero weight. Note the discussion of vertical acceleration, constant velocity, and deceleration in the THK reference.

Dear Jrmichler
I have read what you have written to me and also what is written in the link.
I do not understand
The first answer told me that "In order to solve this problem, show all of the above on one or more diagrams. If a mass is moving in translation, search terms ball screw load inertia will lead to the information to calculate the inertia reflected to the motor.Each acceleration, friction, and gravity force is independent, so you calculate each individually.Then add them up in the final step.

The second time you see I have sent you a doc.pdf where on the second sheet there is a power diagram.
I would be grateful if you could (I would like you to correct what is written if necessary) indicate if the weight that I calculate (written in red) is the weight that must be added to the power screw to calculate the total moment of inertia?
It is square in blue

#### Attachments

• 262.8 KB Views: 23
jrmichler
Mentor
If I understand correctly:
A motor is directly connected to a ball screw.
The ball screw is supported by two bearings.
The ball screw has a ball nut.
The ball nut is connected to a cross shaft.
The cross shaft has a bearing and a weight on each end.
Each cross shaft bearing rides on a surface.

Is the above correct? If so, your weight W should be the total moving mass of the ball nut, the cross shaft, the bearings on the cross shaft, and the weights on the cross shaft.

It is helpful to sketch the motion profile. This is a set of curves showing acceleration, velocity, and position vs time. Use a common time axis, and offset each curve vertically with separate vertical axes. A hand sketch is normally good enough.

When working on these types of problems, it helps to do two or three separate sets of calculations:
1) Inertia, acceleration, and inertial forces and torques.
2) Friction forces and torques.
3) (If applicable) Gravity forces and torques.

For the first set, inertia, here's how I do it (and I have done many of these calculations):
List each inertia separately (all inertia values are reflected to the motor):
Motor
Coupling
Ball screw
The bearings supporting the ball screw
Rotational inertia due to the load

Do the same for the friction forces and torques, but do it on a separate sheet. the last step is acceleration (radians per second^2) times inertia to get torque. Then add the friction torque. Remember to check your units.

And please be aware of significant digits. There is effectively no difference between 200.0000 RPM and 200.0001 RPM, or even 200.0 RPM and 200.1 RPM.

• Lnewqban
If I understand correctly:
A motor is directly connected to a ball screw.
The ball screw is supported by two bearings.
The ball screw has a ball nut.
The ball nut is connected to a cross shaft.
The cross shaft has a bearing and a weight on each end.
Each cross shaft bearing rides on a surface.

Is the above correct? If so, your weight W should be the total moving mass of the ball nut, the cross shaft, the bearings on the cross shaft, and the weights on the cross shaft.

It is helpful to sketch the motion profile. This is a set of curves showing acceleration, velocity, and position vs time. Use a common time axis, and offset each curve vertically with separate vertical axes. A hand sketch is normally good enough.

When working on these types of problems, it helps to do two or three separate sets of calculations:
1) Inertia, acceleration, and inertial forces and torques.
2) Friction forces and torques.
3) (If applicable) Gravity forces and torques.

For the first set, inertia, here's how I do it (and I have done many of these calculations):
List each inertia separately (all inertia values are reflected to the motor):
Motor
Coupling
Ball screw
The bearings supporting the ball screw
Rotational inertia due to the load

Do the same for the friction forces and torques, but do it on a separate sheet. the last step is acceleration (radians per second^2) times inertia to get torque. Then add the friction torque. Remember to check your units.

And please be aware of significant digits. There is effectively no difference between 200.0000 RPM and 200.0001 RPM, or even 200.0 RPM and 200.1 RPM.
Thank you very much

Hi to everyone
I would like if someone has a mathematical "guide" to how a DC motor is chosen, the motor has a gear attached. In the mathematical formulas the parameters of the DC motor are reflected
as, torque constant, voltage constant, motor NL current, motor terminal resistance, ext '. The system parameters I have are:
accelaration moment Tacc 0.0191 Nm
deccelaration moment Tdcc = -Tacc 0.0188 Nm
gravity moment Tg 0.0000 Nm
moment Trms Trms 0.0184 Nm
angular velocity ωPK 20.9440 rad / sec
angular velocity ωPK 200.0000 rad / min
max. power Ppk (motor) 0.40 w
average power Pavg (motor) 0.94 w
screw lead Lsc 0.0127 m / rev
Drive power supply Vdc 12 Vdc
maximum current Ipk-ps Tacc A
Corrent RMS Irms- PS Tdcc A

Note: Merged with earlier thread by moderator

#### Attachments

• 254.5 KB Views: 24
Last edited by a moderator:
jim mcnamara
Mentor
Nice question. For clarity: the weight, 5Kg, is split into two separate 2.5Kg masses, correct?

jrmichler
Mentor
First things first:

angular velocity ωPK 20.9440 rad / sec
angular velocity ωPK 200.0000 rad / min
Are these angular velocities at two different points? If so, you need to say where they are. If they are at the same point, at least one of them is wrong because 20.9 rad/sec is not equal to 200 rad/min. It is best to use one set of units, rad/sec is preferred.

max. power Ppk (motor) 0.40 w
average power Pavg (motor) 0.94 w
Is this the power that the motor needs to deliver? Peak power is always greater than average power.

Motor selection is a lot less confusing if you focus on the following:
1) Peak motor RPM required.
2) Peak motor torque required.
3) Average motor torque required.
Calculate these three, then look for a motor that will deliver at least that much.

On a low speed system such as yours, the friction torque will be larger than your calculations show, so make sure you select a motor accordingly.