Calculate Centre of Gravity/Mass of Non-Uniform Objects Using Load Cells

Click For Summary

Discussion Overview

The discussion revolves around calculating the center of gravity or mass of a non-uniform object using load cells positioned at varying points. Participants explore methods to achieve this without complex calculus, focusing on both two-dimensional and three-dimensional approaches.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant seeks guidance on calculating the center of gravity using load cells, emphasizing the need for a solution that avoids complex calculus due to programming constraints.
  • Another participant suggests that if the positions of the three points and their weight readings are known, the center of gravity can be calculated using simple arithmetic for a two-dimensional case, while noting that a three-dimensional calculation would require additional measurements.
  • A third participant confirms the three-dimensional nature of the problem and expresses confusion regarding the calculations needed, referencing a Wikipedia article but indicating difficulty in understanding the process.
  • One participant provides a formula for calculating the center of mass in two dimensions, suggesting that repeated measurements can yield lines that intersect to determine the three-dimensional center of mass.
  • Another participant shares a link to their work but expresses uncertainty about how to derive coordinates from their calculations, seeking clarification on their approach.
  • Participants discuss the importance of maintaining consistent weight measurements and the implications of rotating the object between measurements, with some confusion about how to represent the results graphically.
  • There is a repeated inquiry about how to draw a line with only one coordinate, highlighting a misunderstanding of the dimensionality involved in the calculations.

Areas of Agreement / Disagreement

Participants generally agree on the need for multiple measurements to determine the center of gravity in three dimensions, but there is confusion and disagreement regarding the specifics of the calculations and how to represent the results graphically.

Contextual Notes

Participants express uncertainty about the calculations and the graphical representation of the center of gravity, indicating potential limitations in their understanding of the mathematical principles involved.

cps.13
Messages
40
Reaction score
1
Hi,

I need to calculate the centre of gravity/mass of a non uniform object. The setup will be using load cells to measure the weight of the object in moveable positions.

If you imagine an object being lowered onto 3 points, the weight reading at each point can be measured individually. But these points will not always be in the same position (i.e. I will not have a permanent equilateral triangle of my 3 points).

I am hoping for an answer in co-ordinates. I can pin point a datum point if required.

Can anyone point me in the right direction? I would like to avoid a calculus method as I will need to program it into a PLC which does not have complex calculus instructions built in.

I thought about doing it using vectors but I cannot figure out how.

Many thanks,
 
Physics news on Phys.org
Is this a three-dimensional problem or just two-dimensional?
If you know the positions of the three points and their weight readings, you can assume your object just has three mass points at the corresponding points, and calculate the (2D) center of gravity just based on those three masses (just additions and divisions, nothing complex). If you need the 3D center of gravity, you'll have to repeat the measurement with a different orientation of the object, and find the intersection of two lines. This does not need complex calculations either.
 
it will be a three dimensional object. Weighing it twice isn't a problem. what calculations do you need to do?

I found this http://en.wikipedia.org/wiki/Center_of_mass under the "locating the centre of mass" section but I got lost with the calulation.

Thanks,
 
Assuming your forces are all purely vertical, you can simplify the third equation to
$$\vec{R}=\frac{1}{W} \left( F_1 \vec{r_1} + F_2 \vec{r_2} + F_3 \vec{r_3} \right)$$
That gives you the 2-dimensional center of mass, or equivalently one line through the object that goes through the 3D center of mass. Define some coordinate system, express that line in that coordinate system, repeat with another measurement, look for the intersection of the two lines.
 
Ok, so far this is what I have done...

https://www.dropbox.com/s/j712tkvuaul9rk0/graphs.jpg?dl=0

I have done (I think!) the calculation you have said to do, but I don't understand how these will provide you co-ordinates to draw a line. I only get one set of co-ordinates.

Are you able to tell me where I'm going wrong?

Thanks,
 
The third dimension gives the line.
How did you rotate your object between the measurements?

The total weight should not change...
 
I know the weight shouldn't change. This is an issue I need to address with the load cells but not my primary concern at the moment. I meant to state that in my last post but forgot!

What do you mean the third line? at present I can only draw one line. Am I getting my first two coordinates correct?

Thanks.
 
cps.13 said:
at present I can only draw one line
Both measurements give you one line each.
cps.13 said:
Am I getting my first two coordinates correct?
The approach is right, I didn't check the numbers.
 
But how do you draw one line with one coordinate?

Thanks
 
  • #10
cps.13 said:
But how do you draw one line with one coordinate?
Two coordinates (x and y) in the horizontal plane allow you to draw a vertical line in three dimensional space.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 31 ·
2
Replies
31
Views
3K
  • · Replies 13 ·
Replies
13
Views
6K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 52 ·
2
Replies
52
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 10 ·
Replies
10
Views
6K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K