# How to Determine the Life of a Linear Actuator Based on Given Specifications?

• alzabroo
In summary: Once you have these values, you can plug them into Eq. 3 to calculate the Equivalent Speed (Vm). This value can then be used in Eq. 4 to calculate the modified actuator life in hours (L10 (hrs) var).In summary, to solve this problem, you will need to determine the values of Cr and Pr for the specific actuator being used, use Eq. 1 to calculate L10, use Eq. 2 to calculate the rotation speed (n), use Eq. 3 to calculate the Equivalent Speed (Vm), and finally use Eq. 4 to calculate the modified actuator life in hours (L10 (hrs) var).
alzabroo

## Homework Statement

The problem deals with a linear rail system not unlike the one displayed in the image below

The relevant variables include:
Load speed required = 2.3 m/s
Cycle time = 9 sec
Capacity of actuator = 33,000 N

With the given information:
A. determine the values for the graph below

B. ...and the life of the actuator in meters
C. ...and the life of the actuator in hours

## Homework Equations

Eq. 1

Where:
L10= actuator lifetime (in millions of revolutions)
Eq. 2

Where:
n = rotation speed (min-1)
this second equation results in unit of hours
Eq. 3
Vm = (V1q1+V2q2+...+Vnqn)/100%
Where:
V1,V2,...Vn = speed at phases of motion profile (in m/min)
q1,q2,...qn = % of time for each phase

Eq. 4
L10 (hrs) var=(L10*105)/(60*Vm)
Where:
L10 (hrs) var = actuator life in hours modified for variable speed
Vm = equivalent speed

## The Attempt at a Solution

There are a few assumptions I've made in my solution attempt to this problem, but I don't know where else to start.

First I assume that the dynamic load rating (Cr) is synonymous with the load capacity of the actuator
Second I assume that the Equivalent Load factor (Pr) is synonymous with the load provided (because no details of load distribution were provided)

To determine the life of the actuator in (millions of revolutions) I input the given values into Eq. 1:

L10=(33,000N/311.5N)3=1.2*106

Eq. 2:

L10 (hrs)=(106/(60*n))*(L10)=(2.0*1010)/n

I'm very unsure of how to determine n, but since the cycle time is 10 sec:
n = 6(min-1) ?

Assuming this, then:
L10 (hrs)=3.3*(109)

In order to use Eq. 3 account for the variable speed of the actuator, I believe I need to determine the values (time increments) of the graph given, but I'm unclear on how to do this with the given info.

I realize this a very long post, but I wanted to be as clear as possible.
Any pointers on how to where I have gone wrong in my process and/or how to proceed would greatly appreciated.
Thank you

Thank you for bringing this problem to our attention. It seems that you have made some good progress in your attempt to solve it. However, there are a few things that need to be clarified in order to provide a complete and accurate solution.

Firstly, it is important to note that the dynamic load rating (Cr) and the load capacity of the actuator are not necessarily the same thing. The dynamic load rating is a measure of the maximum load that the actuator can handle without failure, while the load capacity is the actual load being applied in the system. These two values may be different depending on the specifications of the actuator.

Secondly, the Equivalent Load Factor (Pr) is not the same as the load provided. It is a value that takes into account the type of load being applied (e.g. static, dynamic, shock) and the type of motion (e.g. continuous, intermittent) in order to calculate the equivalent load on the actuator. This value is typically provided by the manufacturer and is specific to the actuator being used.

In order to calculate the life of the actuator in (millions of revolutions), you will need to know the actual values of Cr and Pr for the specific actuator being used. These values can then be plugged into Eq. 1 to calculate L10.

To determine the rotation speed (n), you will need to use Eq. 2 and solve for n. This will give you the rotation speed in revolutions per minute (min-1). You can then use this value to calculate the life of the actuator in hours using Eq. 3 and 4.

To use Eq. 3, you will need to determine the values (time increments) of the graph given. This can be done by breaking down the motion profile into different phases and determining the speed and time for each phase. For example, if the actuator starts at 0 m/min, then accelerates to 2.3 m/s over a period of 2 seconds, then maintains a constant speed of 2.3 m/s for 5 seconds, and finally decelerates to 0 m/min over a period of 2 seconds, then the values for Eq. 3 would be:

V1 = 0 m/min, q1 = 2 sec
V2 = 2.3 m/s, q2 = 5 sec
V3 =

## 1. What is a linear actuator motion profile?

A linear actuator motion profile is a description of how a linear actuator moves over time. It includes information about the speed, acceleration, and position of the actuator at different points in time.

## 2. How is a linear actuator motion profile created?

A linear actuator motion profile is typically created using mathematical equations or computer programming. It takes into account the desired movement of the actuator, as well as any external factors such as load or friction, to determine the necessary speed and acceleration at each point in time.

## 3. What are the different types of linear actuator motion profiles?

There are several types of linear actuator motion profiles, including constant velocity, ramping, and sinusoidal. Each type has its own purpose and is used in different applications depending on the desired movement and precision required.

## 4. Can a linear actuator motion profile be customized?

Yes, a linear actuator motion profile can be customized to fit specific needs. The speed, acceleration, and position can be adjusted to achieve the desired movement for a particular application or task.

## 5. How does a linear actuator motion profile affect the performance of a system?

A well-designed linear actuator motion profile can greatly improve the performance of a system. It can ensure smooth and precise movement, reduce wear and tear on the actuator, and increase overall efficiency and productivity.

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