Jeff, I have been concentrating on the errors for the rectangular rings .... and I am thinking your figure is referring to the CoG locational errors for the triangular rings.
If the size of "dx" is cut in half then the CoG locational errors for the triangular rings is indeed cut in half and the quantity of the triangular rings doubles ... but there are CoG locational errors for each of the rectangular rings to consider, also.
My diagram shows ALL the errors at each iteration
when the cone is analyzed the way you are doing it.
Except that, apparently, instead of using the distance
you want, from P to the center of the rectangle, you
are using the irrelevant distance to the corner of the
rectangle. Which is the error you are obsessing over.
No there are not. The center of the rectangle is the
place you measure to, not the corner.
-- Jeff, in Minneapolis
http://www.FreeMars.org/jeff/
"I find astronomy very interesting, but I wouldn't if I thought we
were just going to sit here and look." -- "Van Rijn"
"The other planets? Well, they just happen to be there, but the
point of rockets is to explore them!" -- Kai Yeves
But why? It just seems to be confusing you. Surely, x_{0} is before x_{1}, why not use (x_{0}+x_{1})/2, like you want to?
Regardless, you seem to have skipped over my point about percentages, and their limit--that, I think, is a more important point. If you understand what happens in the limit, it doesn't matter where you place your first x (but, imagining where you think it is most appropriate should help with your mental feel for the problem.)
http://www.FreeMars.org/jeff/
"I find astronomy very interesting, but I wouldn't if I thought we
were just going to sit here and look." -- "Van Rijn"
"The other planets? Well, they just happen to be there, but the
point of rockets is to explore them!" -- Kai Yeves
27 if you only count once the cut edges which the cubes
have in common.
-- Jeff, in Minneapolis
http://www.FreeMars.org/jeff/
"I find astronomy very interesting, but I wouldn't if I thought we
were just going to sit here and look." -- "Van Rijn"
"The other planets? Well, they just happen to be there, but the
point of rockets is to explore them!" -- Kai Yeves
For the large cube the location of the CoG is sqrt[(x/2)^{2}+(y/2)^{2}+(z/2)^{2}] = sqrt(x^{2}+y^{2}+z^{2})/2
When the cube is diced up into 8 equally-sized cubes the individual CoGs are located at sqrt[(x/4)^{2}+(y/4)^{2}+(z/4)^{2}] = sqrt(x^{2}+y^{2}+z^{2})/4
The size of the edges decreases by a factor of 2 and the location of the CoGs also decrease by a factor of 2 .... but now there are 8 individual CoGs. The CoG is tied to the volume not the edges.
I think we both agree the CoG should be in the center of the rings (technically the y-component is not in the center ... but we simplify).
The reason I keep emphasizing the upper right corner is because that is where the double-integral places the CoG. For the double-integral there is a distance calculated:
Distance:
It places the CoG at the coordinate of (x,y) but what is the relationship of that coordinate with respect to the rectangular shape defined by dx & dy? My pictures in post 276 and post 189 are my attempts to illustrate this.
If the double-integral places the CoG in the wrong place then how can we expect it to produce a correct answer?
Exactly my point. The integral calculates gravity, not the length of your red lines. The total length of your red lines has absolutely nothing to do with the total error.
Which, btw, is zero.As I asked before, why not change your pictures, so that you are better able to understand the integral?
Because, in the limit, the error is zero.If the double-integral places the CoG in the wrong place then how can we expect it to produce a correct answer?
The red lines are just a graphical representation of the misplacement of the CoGs.
I know that the integral does not calculate the length of the red lines.
The integral calculates the distance which is supposed to be as shown in post 189 but instead the double-integral calculates the distance to the upper right corner of the rings.
Because the double-integral is set up incorrectly then every ring has the wrong distance and the wrong "x" for the cosine calculation.
After your cube demonstration I realized that the number of errors in CoG location is actually a cubic function not a squared function:
Quantity of Errors = n^{3}+n*T_{n}
The integral calculates a "dM" which I talked about in post 233 and each of these "dM" chunks has an erroneous location for its CoG.
But I think the "ring" concept captures that 3rd dimension so we can just stick with
Quantity of Errors = n^{2}+T_{n}
Absolutely wrong. It calculates force on a unit test particle, not distance.
It doesn't matter what the number of errors is--in fact, it could be infinite--so long as the limit of their overall effect is zero. Which it is in this case.Because the double-integral is set up incorrectly then every ring has the prong distance and the wrong "x" for the cosine calculation.
After your cube demonstration I realized that the number of errors in CoG location is actually a cubic function not a squared function:
Quantity of Errors = n^{3}+n*T_{n}
grapes,
Ignoring the fact that it makes no difference to the result
of the integration, how can xylophobe modify his technique
so that his calculation refers to the center, not the corner
of each rectangle?
-- Jeff, in Minneapolis
http://www.FreeMars.org/jeff/
"I find astronomy very interesting, but I wouldn't if I thought we
were just going to sit here and look." -- "Van Rijn"
"The other planets? Well, they just happen to be there, but the
point of rockets is to explore them!" -- Kai Yeves
I'm mostly showing that both you and I arrive at the same conclusion which is shown in the picture from post 189.
The x-coordinate should be x-dx/2
The y-coordinate should be y-dy/2 (simplified version)
I am aware that you know that, but you still said what you did.
The difference is, what I said earlier does not contradict what I said later.but the sub-steps of the whole calculation are a distance, a cosine, and a volume.
You even state the same (underlined and highlighted in red) in your post below:
Then you've missed my point.
Try using an x-coordinate of x (and a y-coordinate of y) to the center of the rectangle. That should clarify a lot of what is going on.
http://www.FreeMars.org/jeff/
"I find astronomy very interesting, but I wouldn't if I thought we
were just going to sit here and look." -- "Van Rijn"
"The other planets? Well, they just happen to be there, but the
point of rockets is to explore them!" -- Kai Yeves
If the double-integral functions as has been described to me then the following integrals should all produce the same answer:
Upper Right Corner
Lower Right Corner
Center
Lower Left Corner
Upper Left Corner
because the difference in the location of the CoGs would disappear as the size of "dx" approached zero.
The above represent the four corners of the rectangle defined by "dx" & "dy" and also the center of the rectangle.
Last edited by xylophobe; 2012-May-23 at 03:21 PM.
It may not make any real difference, but I would expect
the correct function for the center to have the smallest
number of factors, not the largest. Rather than dividing
by two, I think you should re-define something.
-- Jeff, in Minneapolis
http://www.FreeMars.org/jeff/
"I find astronomy very interesting, but I wouldn't if I thought we
were just going to sit here and look." -- "Van Rijn"
"The other planets? Well, they just happen to be there, but the
point of rockets is to explore them!" -- Kai Yeves
That was my mistake .... I just forgot to change the denominators (I fixed it).
I finally figured out how to check whether the five (5) different double-integrals will produce the same answer by using numerical integration. I had been trying to use an excel spreadsheet but by the time I reached 32 divisions of the interval "h" it was becoming quite cumbersome .... so I wrote a VBA program (below). The visual basic program below only compares the upper right corner CoG with the centered CoG and the answer is the same as produced by:
So now I can confirm that the double-integral, as originally written, will produce the correct result (although I have not checked one cone segment with another). Using numerical integration and the upper right corner of the double-integral it would take my computer 35,945 years to calculate the result accurately to 9 significant digits; whereas, using the centered CoG the calculation time is only 4.7 hours to get the result accurately to 10 significant digits.
Sub Calc1()
Dim StartTime As Double, EndTime As Double
StartTime = Timer
h = Range("G12").Value
r = Range("G13").Value
n = Range("G14").Value
i = 0
j = 0
SumOf0 = 0
SumOf1 = 0
Pi = Range("G15").Value
Nloop:
i = i + 1
If i > n Then GoTo ExitLoop Else
j = 0
j2 = 0
SubLoop:
j = j + 1
If j > n + i Then GoTo SubLoop2 Else
Distance0 = Sqr((h + i * h / n) ^ 2 + (j * r / n) ^ 2)
InverseSquare0 = 1 / Distance0 / Distance0
Cosine0 = (h + i * h / n) / Distance0
Mass0 = 2 * Pi * (h / n) * (r / n) * (j * r / n)
SumOf0 = SumOf0 + Cosine0 * InverseSquare0 * Mass0
GoTo SubLoop
SubLoop2:
j2 = j2 + 1
If j2 > n + i - 1 Then GoTo SubLoop3 Else
Distance1 = Sqr((h + i * h / n - 0.5 * h / n) ^ 2 + (j2 * r / n - 0.5 * r / n) ^ 2)
InverseSquare1 = 1 / Distance1 / Distance1
Cosine1 = (h + i * h / n - 0.5 * h / n) / Distance1
Mass1 = 2 * Pi * (h / n) * (r / n) * (j2 * r / n - 0.5 * r / n)
SumOf1 = SumOf1 + Cosine1 * InverseSquare1 * Mass1
GoTo SubLoop2
SubLoop3:
Distance11 = Sqr((h + i * h / n - 1 * h / n / 3) ^ 2 + (j2 * r / n - 2 * r / n / 3) ^ 2)
InverseSquare11 = 1 / Distance11 / Distance11
Cosine11 = (h + i * h / n - 1 * h / n / 3) / Distance11
Mass11 = 2 * Pi * 0.5 * (h / n) * (r / n) * (j2 * r / n - 2 * r / n / 3)
SumOf1 = SumOf1 + Cosine11 * InverseSquare11 * Mass11
GoTo Nloop
ExitLoop:
EndTime = Timer
Range("G16") = SumOf0
Range("G17") = SumOf1
Range("H13") = EndTime - StartTime
End Sub
Last edited by xylophobe; 2012-May-23 at 05:14 PM.
If that is giving you the answer you expected, what happened to the "twice"?
When I started writing the program I just did it for the cylindrical (rectangular) portion (of segment 2) but the mass of the triangle part of the frustum equals 4/3 of that of the cylindrical part so I was missing over half of the mass on my first go around.
It would be good of you if you could publish the step-by-step solution to the double-integral. I can not figure the correct way to deal with the denominator.
Thanks.
There is no correct way, but some ways are easier than others. My approach from post#194:Except, now that I look at it, the inside integral is the y-integral, so I'd want to use A dA = y dy, then that integral is just dA/A^(2) times some constants.
If you use the second substitution, the integral becomes dA/A^(3/2), unless I've made *another* mistake this morning.