Cones - Center of Gravity, Center of Mass, etc.

[ShinAce]

Any shape can be hollow or solid. This thread is about solid objects.

Don't take the pizzas too literally. We start with a cone and slice it up. The walls of that pizza are not parallel. They are approximately parallel for thin pizzas. You can then slice the pizza into triangles. As long as each triangle has the same surface area, it will have the same weight. Weight is not mass. Weight is a force. We want to calculate forces, so the gravitational force is weight.

A cylinder is anything with parallel walls, which we don't have. The majority of pools, including kidney shaped pools are cylinders. The base can be any closed 2d surface. But cylinders don't reduce to a point at the origin.

A cone is any closed surface made by pivoting a line through the origin. Mathematically speaking, a pyramid is a type of cone. But you would integrate a straight sided pyramid in cartesian coordinates and elliptical based cones in cylindrical coordinates.

The wiki page for cones covers their definition.

[xylophobe]

First, thanks for the explanation of the double-integral ... I'll probably have questions as is I think about it.

... 2pi y is the circumference of a ring which when multiplied by its cross-section dx dy gives the volume of an infinitesimal ring ...

This statement indicates that the formula is using the Pappus calculation to calculate the volume which if done correctly then the "y" should be y/2 because the centroid of the cross-sectional area, which is a rectangle (with y for height), is half the height value (the x-loc is also half the x-length but that is not used in the Pappus volume calculation).

If the integration takes equal horizontal slices then the centroid for the 5th slice is located at 4*dy + dy/2 which is the "r" for the area [dx*dy] times the distance the centroid travels [circumference = 2*pi()*r] around the axis which equals 2*pi()*(4*dy + dy/2)

Centroid of Rectangle (see slide #4 of the presentation)

[grapes]

No, the height of the rectangle is not y, the rectangle is dx wide by dy high, y is essentially the distance to the centroid.

[HUb']

No, the height of the rectangle is not y, the rectangle is dx wide by dy high, y is essentially the distance to the centroid.

On that note, i will just say " i have no idea,,,, this was in January ? it is now April ?/? "
& travel elsewhere: what i would like to find, is an explanation of perigee.bas
which computes the days when the moon reaches PERIGEE & apogee also,
I do have a version of that sky & telescope program from years gone by
i could upload it {Maybe) it is rather long? & maybe i already did? it begins
REM CONSTANTS
DIM A(30), B(30), R(30), A1(30)
DIM C0(30), C1(30), C2(30), C3(30)
DIM C4(30):

THUS: there are 30 parameters ?{objects) of
:Line " 253 DATA 51818,0,1,0,0,0 " of six parameters each
My thought {at this time)i would like to add A 31'st {oh never mind | time is short
AUTO SAVE in effect | Link to where i went to later {Hasta Lego ToV

[xylophobe]

No, the height of the rectangle is not y, the rectangle is dx wide by dy high, y is essentially the distance to the centroid.

Does the integral account for the exponential increase in mass due to increasing radius. The exponential increase in the below picture is only shown in the cylindrical portion of the frustum but it continues into the triangular section which shifts the CoG towards the larger end.

[Jeff Root]

xylophobe,

I'm glad if you are learning from this (it's your thread,
so that's the first priority), but I just keep getting loster
and loster. Maybe HUb' can explain it to me.

Your diagram showing "exponentially increasing mass"
makes absolutely no sense to me. The density of the
hatching seems to be an indicator of material density.
But you have given no reason for the density to vary.
And you show it increasing radially outward from the
centerline of the cone. What on Earth is that about???
Am I completely misinterpreting the diagram and what
you are saying about increasing mass? I would think
the "increasing mass" would be the increase in mass
of each succeeding frustum with increasing distance
from the origin, due to its larger volume. No?

-- Jeff, in Minneapolis

[grapes]

Does the integral account for...

It accounts for everything I can think of.

You can start off with a simple integral that calculates the volume by adding up every single little box,



Then you just multiply that by density to get mass, then multiply by the gravitational constant G and divide by the distance squared to get acceleration, then by the cosine of the angle to get the component parallel to the axis. I think that's everything.

[xylophobe]

xylophobe,

I'm glad if you are learning from this (it's your thread,
so that's the first priority), but I just keep getting loster
and loster. Maybe HUb' can explain it to me.

Your diagram showing "exponentially increasing mass"
makes absolutely no sense to me. The density of the
hatching seems to be an indicator of material density.
But you have given no reason for the density to vary.
And you show it increasing radially outward from the
centerline of the cone. What on Earth is that about???
Am I completely misinterpreting the diagram and what
you are saying about increasing mass? I would think
the "increasing mass" would be the increase in mass
of each succeeding frustum with increasing distance
from the origin, due to its larger volume. No?

-- Jeff, in Minneapolis

I guess some of my pictures only make sense to me so I'll try to explain.

In that diagram the density is constant ... the cross-hatching was not intended to imply a changing density.

The mass increases exponentially because the mass of a cylinder is pi()*r²h so as "r" increases the mass increases according to the square of "r".

My increments were not exactly precise (I realized after I posted) because the increments were supposed to indicate rings of equal mass.

Let me know if you have additional questions.

[xylophobe]

It accounts for everything I can think of.

You can start off with a simple integral that calculates the volume by adding up every single little box,



Then you just multiply that by density to get mass, then multiply by the gravitational constant G and divide by the distance squared to get acceleration, then by the cosine of the angle to get the component parallel to the axis. I think that's everything.

Cut it out .... I'm still trying to figure out the double-integral ..... jk.

After thinking about it a while I think the picture below is an accurate description of what the double-integral is intended to accomplish.

I only show a single slice along the x-axis and a finite number of slices along the y-axis

[grapes]

I dunno if that really depicts what it is trying to accomplish, but it looks like things are labeled correctly.

[xylophobe]

In the picture below the image on the far left is a cylinder divided into 8 sections with equal radial thickness. The numbers are the masses of each ring, using the innermost section (which is a cylinder) as the unit = 1.0

2 times "r" causes the mass to grow by a factor of 4 which is what 1+3 equals
3 times "r" causes the mass to grow by a factor of 9 which is what 1+3+5 equals
4 times "r" causes the mass to grow by a factor of 16 which is what 1+3+5+7 equals
5 times "r" causes the mass to grow by a factor of 25 which is what 1+3+5+7+9 equals
6 times "r" causes the mass to grow by a factor of 36 which is what 1+3+5+7+9+11 equals
7 times "r" causes the mass to grow by a factor of 49 which is what 1+3+5+7+9+11+13 equals
8 times "r" causes the mass to grow by a factor of 64 which is what 1+3+5+7+9+11+13+15 equals
etc.
etc.
etc.

The 5 images to the right of the cylinder are frustum segments that show how the mass of the conic section varies between a beginning at a 1/3 to 2/3 split towards an infinite limit of 1/2 to 1/2 split. By evaluating the cone segments as infinitely small cylinders the double-integral approaches an exact value the farther from point "P' that the segments are but up close the double-integral is inaccurate.

The chart at the bottom of the picture shows how the two conic sections approach equality because as "x" approaches infinity then x+1 approaches "x"

This geometric analysis proves that the cone segments do not have equivalent gravitational influence upon point "P"

Nevertheless, the double-integral is a valuable tool for determining real-life applications because we do not encounter cone-shaped gravitating objects in outer space ... just in thought experiments.

[xylophobe]

Here is a better pictorial explanation of how the double-integral errs from an exact solution.

There is always a constant difference between the approximation and the actual geometry --- adding a constant eliminates the possibility for reaching an exact solution.

[Grey]

Here is a better pictorial explanation of how the double-integral errs from an exact solution.

Except that your drawing assumes that dx and dy have finite values, and that you're dividing the cone up into a small finite number of sections. The integral is based on letting the size of the sections go to zero, while simultaneously letting the number of such sections approach infinity. When you do that, the amount of error likewise goes to zero. Indeed, that's the entire point of integral calculus.

[grapes]

This geometric analysis proves that the cone segments do not have equivalent gravitational influence upon point "P"

But you haven't even used a distance from the cone segments to the point "P", which is necessary to calculate the gravitational effect.

Your diagrams do not represent how integrals are calculated. The integrals are exact, not approximations, using the same conditions that you assume--that density is constant, and the shapes are perfect cones.

[xylophobe]

But you haven't even used a distance from the cone segments to the point "P", which is necessary to calculate the gravitational effect. ...

If I add the distance from the cone segments as you suggest then everything is fine for the cylindrical portion of the frustum so I will concentrate on just the conic section of the frustum.

The below picture shows the slice as the double-integral estimates which is just a rectangle and because it is wafer thin then its CoG is essentially the same as the centroid for a rectangular area. The "x" and "y" are located, by the integral, in the center and the height of the wafer is "dy" while the width of the wafer is "dx". The distance from point "P" to the CoG for this rectangular wafer solid is shown as a square-root calculation.

The actual geometry has a triangular shaped wafer and because it is a wafer then its CoG matches the location of the centroid for a triangular area. The distance from point "P" to the CoG for this triangular wafer solid is also shown.

What I show is the very first slice from a cone (or segment 1 from my previous pictures) which clearly shows that the CoG for the actual solid is in a different location than what is used by the double-integral calculation.

The difference in actual location for the CoG and the integral-approximation approaches zero difference but only when the cone reaches infinite dimensions.

The error adds up as the length of the frustum increases ... with the most error being close to point "P" and near zero error at the farthest distance from point "P"

[xylophobe]

Except that your drawing assumes that dx and dy have finite values, and that you're dividing the cone up into a small finite number of sections. The integral is based on letting the size of the sections go to zero, while simultaneously letting the number of such sections approach infinity. When you do that, the amount of error likewise goes to zero. Indeed, that's the entire point of integral calculus.

An integral funtions as you state but only for continuous functions - the frustum is two functions: a cylinder and a conic section.

[grapes]

The integral is not an approximation. It is an exact calculation--that is the point of using calculus.

ETA:

An integral funtions as you state but only for continuous functions - the frustum is two functions: a cylinder and a conic section.

I'm not sure exactly what you mean by this, as integrals are applied to multiple functions all the time. **We can illustrate by getting rid of the mass and distance factors and just calculate volume.



Evaluating the theta integral gets us:



Evaluating the y integral brings us to:



Which results in , the exact formula for a cone, no matter what h or r.

[ShinAce]

Integration definitely does give exact answers. Numerical methods, like expanding into a series and integrating, has inherent errors. Be advised that your calculator does many things using series, such as the log function. Integration by hand is not numerical, but it is exact.

I also did the integral for cone volume, and got Pi*r^2*h/3. Exactly.

When I did the integral for force due to gravity, I kept the little m in GmM(it's the test particle, or the mass of your scale) and changed big M to density. I got that F=2*Pi*G*m*density*h / sqrt(1 + r^2/h^2). I also got the curious result that when integrating the third integral dx, x was long gone. It didn't matter if I evaluated x from 1 to 2, or from 3.5 to 4.5. Any slice of fixed height for that cone gives the same gravitational force. Also pretty cool is that I got a negative answer! I integrated with the forces acting from the origin to the infinitesimal 'rings' I was integrating. The answer I get is that the force acts on the test particle, not the other way around like I though I had set it up. Thankfully Newton's forces are equal and opposite pairs.

We can rearrange my expression for F to get:
F = 2*Pi*G*m*density*h^2 / sqrt(h^2 + r^2) , but sqrt(h^2 + r^2) is the length from the cone tip to a point on the base's perimeter. Let's call that s, the length of the slope.
F = 2*Pi*G*m*density*h^2 / s

You can derive r from h and s. It's not lost, just built in.

[grapes]

When I did the integral for force due to gravity, I kept the little m in GmM(it's the test particle, or the mass of your scale) and changed big M to density. I got that F=2*Pi*G*m*density*h / sqrt(1 + r^2/h^2). I also got the curious result that when integrating the third integral dx, x was long gone. It didn't matter if I evaluated x from 1 to 2, or from 3.5 to 4.5. Any slice of fixed height for that cone gives the same gravitational force. Also pretty cool is that I got a negative answer! I integrated with the forces acting from the origin to the infinitesimal 'rings' I was integrating. The answer I get is that the force acts on the test particle, not the other way around like I though I had set it up. Thankfully Newton's forces are equal and opposite pairs.

Thanks for re-doing the integral, it's always nice to have a check. However, your answer differs from mine. You shouldn't have gotten a negative result--all constants and variables are positive. Apparently, you left off the second (positive) value of the interior definite integral, which is lessened by that negative value.

We can rearrange my expression for F to get:
F = 2*Pi*G*m*density*h^2 / sqrt(h^2 + r^2) , but sqrt(h^2 + r^2) is the length from the cone tip to a point on the base's perimeter. Let's call that s, the length of the slope.
F = 2*Pi*G*m*density*h^2 / s

Notice, for a constant h, your F decreases as s increases. That can't be right. As s gets larger, the cone still contains the smaller cones--so the force can't decrease.

[ShinAce]

The only place I can think of where I made a mistake would be the infinitesimal masses of the cone, dM. I'll redo it later as a double integral of rings(aka, drop r*dtheta and replace with 2*Pi*r), instead of the triple integral I tried out.

My expression for the integration looked a lot like yours. All I had to do to get the same orientation of the cone was change rdr for ydy. I had a cos(phi) in the expression for F that comes about due to symmetry arguments. That reduces to x/d where d is the distance between the particles acting on each other. That gave me the same:
xy/(x^2 + y^2)^3/2 dphi dy dx
Integrating that with respect to y gave me the negative. Now you can rearrage the denominator to cancel out x entirely. When I integrate with respect to x, I don't get a second negative to cancel out the first one.

[xylophobe]

In the attached picture I show the integration using horizontal slices in the above image and integration using vertical slices in the bottom image.

The integration using vertical slices yields a correct mass because all the "dx" slices are bounded similarly: "x-axis" on the bottom, "x" on the large end, "x-h" on the small end, and "y=xr/h" on the top. The function is continuous along the entire length.

The integration using the horizontal slices is not over a continuous function because there is a discontinuity at the common point between the boundaries "x-h" and "y=xr/h" (the point) . This violates a fundamental rule of calculus. The left-most boundary of the horizontal slices is "x-h" for the cylinder section and "y=xr/h" for the conic section. The center of gravity for the horizontal slices in the conic section is constantly changing for each "dy" slice.

[xylophobe]

Here are the calculations for the CoGs for the frustums. You can imagine them as being the whole frustum or as three (3) individual "dx" slices ... plus the first cone-shaped "dx" slice.

[xylophobe]

Here are the calculations for the CoGs for the frustums. ...

I was just looking at the picture and if I do not increase the viewing scale on my browser then the divide symbol that I used looks like a "+" symbol .... I guess I should have stuck with the old standby "/" symbol. Hopefully this is only happening on my computer and not everyone else's.

[grapes]

That gave me the same:
xy/(x^2 + y^2)^3/2 dphi dy dx
Integrating that with respect to y gave me the negative.

That should result in -x/(x^2 + y^2)^1/2 dx (dphi?), but now you have to evaluate it at y=rx/h *and* subtract it evaluated at y=0

[ShinAce]

*and* subtract it evaluated at y=0

Oops! There it is.

[xylophobe]

The integral is not an approximation. It is an exact calculation--that is the point of using calculus.

ETA: I'm not sure exactly what you mean by this, as integrals are applied to multiple functions all the time. We can illustrate by getting rid of the mass and distance factors and just calculate volume.



Evaluating the theta integral gets us:



Evaluating the y integral brings us to:



Which results in , the exact formula for a cone, no matter what h or r.

These are excellent confirmations that a properly set up integral will yield a correct answer - now change the limits to get just a frustum volume - either segment 2, segment 3, or segment 4.

As far as applying integrals to multiple functions I think what you are describing is multiple variables, ie: , x, y

What I am referring to is multiple functions --- the function of the sloped cone line y=xr/h, the shape of the horizontal cylinder line y=r (around x-axis for segment 1)

[grapes]

Thank you. It wasn't clear that you agreed with the frustrum results, I was kinda under the impression that you were trying to argue against them.

[Grey]

In the attached picture I show the integration using horizontal slices in the above image and integration using vertical slices in the bottom image.

The integration using vertical slices yields a correct mass because all the "dx" slices are bounded similarly: "x-axis" on the bottom, "x" on the large end, "x-h" on the small end, and "y=xr/h" on the top. The function is continuous along the entire length.

The integration using the horizontal slices is not over a continuous function because there is a discontinuity at the common point between the boundaries "x-h" and "y=xr/h" (the point) . This violates a fundamental rule of calculus. The left-most boundary of the horizontal slices is "x-h" for the cylinder section and "y=xr/h" for the conic section. The center of gravity for the horizontal slices in the conic section is constantly changing for each "dy" slice.

It's perfectly possible to use calculus to calculate this in either case. It's true that if you do it using "horizontal slices" as you have it shown here, you'll have to break the integral into two pieces, but that doesn't stop you from using calculus to solve the problem, or from getting an exact answer. As far as I can tell, none of the integrals that grapes has set up would even have this as an issue at all. The integrals he's been using are all the correct method to determine the gravitational force from a cone (or a portion thereof) on a point.

[xylophobe]

It's perfectly possible to use calculus to calculate this in either case. It's true that if you do it using "horizontal slices" as you have it shown here, you'll have to break the integral into two pieces, but that doesn't stop you from using calculus to solve the problem, or from getting an exact answer. As far as I can tell, none of the integrals that grapes has set up would even have this as an issue at all. The integrals he's been using are all the correct method to determine the gravitational force from a cone (or a portion thereof) on a point.

So far I have not seen any integrals that calculate for a frustum. If the integral can not solve for the volume of a frustum then the gravitational force that the integral yields is also not correct for a frustum.

I asked grapes to solve his triple integral for the volume of a frustum because I am sure that he can not because the triple integral is set up to solve for cones not frustums --- I calculated the volumes of the frustums and so far no integral has produced the same result.

Below I will dissect the grapes double-integral into its component parts to show that the volume must match the volume for a frustum.




Constants:


Distance:


Inverse Square of Distance:


Cosine:


Leaves the Volume of a Cone:


This yields:

This double-integral should solve for the volume of the object being examined which for our discussion is a frustum but it does not solve for the volume of a frustum because it results in the volume of a cone --- I have already shown (post 145) that cones follow all the claims: mass squares with distance which cancels out the inverse square of the distance.

This double-integral will not solve for the volume of a frustum because a frustum involves two functions: a cylinder and a conic section. To get the correct solution for a frustum the solver would have to solve individually for the cylinder and then for the conic and then add the two component solutions to receive the correct total solution.

Violating the fundamental requirement (continuity) of integrals results in an integral that yields an incorrect solution --- fix the integral and it will solve for what we are looking for!!

btw: what is the code for creating the brackets around an integral solution, with the limits? I need this to show how the integral is solved. Thanks.

[grapes]

The (interior) y integral stays the same, we just change the limits on x

Evaluating the y integral brings us to:



Which results in , the exact formula for a cone, no matter what h or r.

So, for a frustrum which has half the height of the cone, the limits would go from h/2 to h, thereby cutting off the top part of the cone:


That becomes the evaluation:



Hmmm, maybe this looks better:



Which is the same as for the cone, except for the different limits. It evaluates to:



Which is of course 7/8 of the original cone, same answer you got.