# Centre of gravity

1. Feb 12, 2015

### cps.13

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,

2. Feb 12, 2015

### Staff: Mentor

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.

3. Feb 12, 2015

### cps.13

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,

4. Feb 12, 2015

### Staff: Mentor

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.

5. Feb 13, 2015

### cps.13

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,

6. Feb 13, 2015

### Staff: Mentor

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

The total weight should not change...

7. Feb 13, 2015

### cps.13

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.

8. Feb 13, 2015

### Staff: Mentor

Both measurements give you one line each.
The approach is right, I didn't check the numbers.

9. Feb 14, 2015

### cps.13

But how do you draw one line with one coordinate?

Thanks

10. Feb 14, 2015

### jbriggs444

Two coordinates (x and y) in the horizontal plane allow you to draw a vertical line in three dimensional space.