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## Center of gravity and center of ???  24-Ruby III

## Center of gravity and center of ???

I try to calculate the center of gravity one not simple plane figure.

What is it the red point? Center of mass

1 ACCEPTED SOLUTION

Accepted Solutions  18-Opal
(To:ValeryOchkov)

The red point is the geometric median‌.

45 REPLIES 45  24-Ruby III
(To:ValeryOchkov)

3D variant   23-Emerald I
(To:ValeryOchkov)

Where did you learn that the center of gravity was y = 0.5?  18-Opal
(To:Fred_Kohlhepp)

y:=0.5m was just his guess value for the subsequent Minimize function.

We are all in agreement that the center of mass is at y=61cm.

What is the red point?  As defined by the equation, it is the point with the shortest distance to the mass.  If it has a name, I don't know it.  23-Emerald I
(To:MJG)

But when I do the same thing: So I don't see where 74.883 came from.  18-Opal
(To:Fred_Kohlhepp)

Fred,

Your summation should go from i = 0 to n1 (not n1-1); however, this will make a negligible difference.  You would want n1-1 if you had defined n1 with the rows() function, but you used last(), so the -1 is not needed.

The difference in the calculations is that Valery took the square root before summing.  When I do the same in your worksheet, I get the same 74.883 that Valery got.  23-Emerald I
(To:MJG)

So do I!

Why is it wrong?  Minimizing X^2 and minimizing X should give the same result, shouldn't it?  18-Opal
(To:Fred_Kohlhepp)

That's what I thought at first, but we're not minimizing X or X^2.  We're minimizing a sum.

Here's a quick graph: The points near y=0 are furthest from x=0, so their effect is stronger on S1 than they are on S2.  23-Emerald I
(To:MJG)

I think you're right!

I took histograms of X and Y: More of the Y's are near the top, while most of the X's are  at he ends.

Note that the mean of Y gives the same answer as S2 (aka Sq):    18-Opal
(To:Fred_Kohlhepp)

Here's a simplified version of the problem worth pondering:   24-Ruby III
(To:MJG)
 Mark Gase написал(а): Here's a simplified version of the problem worth pondering: One more example   23-Emerald III
(To:ValeryOchkov)

I find it ever so sad to see that you generate a lot of data, then throw away the largest part of it. So that you keep your n points.

Luc  24-Ruby III
(To:LucMeekes)
 LucMeekes написал(а): I find it ever so sad to see that you generate a lot of data, then throw away the largest part of it. How about: So that you keep your n points. Luc

Sorry, but better to have same density of points in horizontal and vertical dimensions!

Not 0..1 and 0..1/2, but 0..1 and 0..1  23-Emerald III
(To:ValeryOchkov)

Does this help?   If you change the 'thickness' of the semi-circle, the amount of points with y above y1 doesn't change as much as y1.

Extreme example: Luc  18-Opal
(To:LucMeekes)

Luc,

The way you have defined the data points, there are more near the inside radius than the outside radius.  This is especially visible in your last graph.

Valery's setup made the points uniformly distributed throughout the whole area.

This will skew some of your calculations.  23-Emerald III
(To:MJG)

Mark,

I agree.

Luc  24-Ruby III
(To:MJG)
 Mark Gase написал(а): Luc, The way you have defined the data points, there are more near the inside radius than the outside radius.

It is not correct but more nice!   23-Emerald III
(To:MJG)

Better like this? Luc  16-Pearl
(To:LucMeekes)

Luc, to get a uniform distribution of points over a circular area you need to generate a uniform distribution for r^2 and then take the square root to get r (as well as generating a uniform distribution for phi, of course).

Alan  23-Emerald III
(To:AlanStevens)

Ah, sure Alan! Why didn't I think of that. So like this: Luc  18-Opal
(To:ValeryOchkov)

The red point is the geometric median‌.  24-Ruby III
(To:MJG)
 Mark Gase написал(а): The red point is the geometric median.

We must check in on one triangle with three medians!  24-Ruby III
(To:ValeryOchkov)

I think the geometric median and the center of gravity must be one point .

Why?

See please two method of the searching this point:  But we have by using our (Monte-Carlo) method;   24-Ruby III
(To:ValeryOchkov)

Sorry!

This point is

Barycenter - Wikipedia

or geometric center.

If our figures have constant density - geometric center = mass center!

Why we have: geometric center and mass center are two different points at constant density?  18-Opal
(To:ValeryOchkov)
 In geometry, the term "barycenter" is synonymous with centroid, the geometric center of a two-dimensional shape.

That means:

barycenter = centroid = geometric center = center of mass w/ constant density

 Valery Ochkov wrote: Why we have: geometric center and mass center are two different points at constant density?

You have two points because the geometric median is not the same as the geometric center.  24-Ruby III
(To:MJG)
 Mark Gase написал(а): In geometry, the term "barycenter" is synonymous with centroid, the geometric center of a two-dimensional shape. That means: barycenter = centroid = geometric center = center of mass w/ constant density Valery Ochkov wrote: Why we have: geometric center and mass center are two different points at constant density? You have two points because the geometric median is not the same as the geometric center.

Sorry, Mark,

but with constant density the geometric median is the same as the geometric center. Or?  24-Ruby III
(To:MJG)

Yes, Mark!

It is the geometric median!

But what is the name this point in German, French, Italian, Russian....   24-Ruby III
(To:ValeryOchkov)

Center of Gravity... 4D-body   24-Ruby III
(To:ValeryOchkov)

Two pictures and sheets for thinking!

Three points! Why? One point!   18-Opal
(To:ValeryOchkov)
 Center of Gravity... 4D-body

Where is the plot?  I want an animated XYZ plot with W as time. 