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

  1. #271
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    Is This Correct Solution

    Quote Originally Posted by grapes View Post
    ...

    The nice thing about the evaluation of that integral is that the y-integral is a constant! The x variable disappears:



    Multiplying by the other four constants ( ) is still a constant, so the effect of the x-integral is just to multiply that resulting constant by the difference in the x limits. In other words, in your cases, h, each time. So, all four segments have the same answer.
    Is this the correct solution for the double-integral?

    So if the 2nd segment has an h=6 and r=1.5 then the answer is 0.029857499854668?

  2. #272
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    Communication

    Quote Originally Posted by grapes View Post
    If that is so, why do you not show it that way on your diagrams?

    I think that would clear up a lot of confusion.
    If I learn nothing else from this thread I will learn how to communicate complicated ideas more effectively .... I hope.

  3. #273
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    Not Sure

    Quote Originally Posted by Jeff Root View Post
    So, why do you use what you know to be incorrect
    locations for all of your calculations? Why not use
    the correct locations? Then the only errors will be
    those I showed in my diagram. Which are clearly
    cut in half at each iteration.

    -- Jeff, in Minneapolis
    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.

  4. #274
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    Quote Originally Posted by xylophobe View Post
    Is this the correct solution for the double-integral?

    So if the 2nd segment has an h=6 and r=1.5 then the answer is 0.029857499854668?
    No, multiply that by 6.
    Quote Originally Posted by xylophobe View Post
    If I learn nothing else from this thread I will learn how to communicate complicated ideas more effectively .... I hope.
    Did you already answer my question somewhere else?

  5. #275
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    Quote Originally Posted by xylophobe View Post
    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.
    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.

    Quote Originally Posted by xylophobe View Post
    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.
    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

  6. #276
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    Quote Originally Posted by grapes View Post
    ... Did you already answer my question somewhere else?
    "dx" is the distance that "x" advances so if the frustum is divided into "n" segments then

    the first "x" = x1 = x0+dx

    and

    dx=(upper limit - lower limit)/n

    Below is a picture that illustrates this.


    Click image for larger version. 

Name:	ConeSegment13.GIF 
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  7. #277
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    Quote Originally Posted by xylophobe View Post
    "dx" is the distance that "x" advances so if the frustum is divided into "n" segments then

    the first "x" = x1 = x0+dxx0
    But why? It just seems to be confusing you. Surely, x0 is before x1, why not use (x0+x1)/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.)

  8. #278
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    Quote Originally Posted by grapes View Post
    For a cube of size 1, the sum of its edges is 12.
    Divide it in half each way (8 smaller cubes), and
    the sum of their edges is 24.
    48.



    -- 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

  9. #279
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    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

  10. #280
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    Quote Originally Posted by Jeff Root View Post
    48.



    -- Jeff, in Minneapolis
    I stand corrected!

  11. #281
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    Location of CoG

    Quote Originally Posted by grapes View Post
    xylophobe, the length of your red lines do increase, but they have very little to do with the calculation itself.

    The same thing can happen with a simple cube. For a cube of size 1, the sum of its edges is 12. Divide it in half each way (8 smaller cubes), and the sum of their edges is 24. Each time you cut it this way, you double the length of the edges total. But the volume stays the same. We have to be careful that what we are measuring is related to what we are trying to find.

    In this particular case, it is not the sum of the red lines that constitute the error. If the maximum length of the lines at a step is R, then the introduced error might be from using x+R instead of just x, a ratio of about R/x error (or, since we're using the square of the distance, the error would be about 2R/x). Each piece might contribute that much error, but it is multiplied by the mass, so the error is small for each piece. If the error were 5% for each piece, the total error would still be only 5%--right?

    That means the total error might be at most 2R/h, since x does not get any smaller than h. As each red line gets smaller, the R becomes smaller. In the limit, there is zero error.
    For the large cube the location of the CoG is sqrt[(x/2)2+(y/2)2+(z/2)2] = sqrt(x2+y2+z2)/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(x2+y2+z2)/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.

  12. #282
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    Location of CoG

    Quote Originally Posted by Jeff Root View Post
    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
    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?

  13. #283
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    Quote Originally Posted by xylophobe View Post
    The CoG is tied to the volume not the edges.
    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.
    Quote Originally Posted by xylophobe View Post
    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.
    As I asked before, why not change your pictures, so that you are better able to understand the integral?
    If the double-integral places the CoG in the wrong place then how can we expect it to produce a correct answer?
    Because, in the limit, the error is zero.

  14. #284
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    x-dx/2

    Quote Originally Posted by grapes View Post
    But why? It just seems to be confusing you. Surely, x0 is before x1, why not use (x0+x1)/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.)
    I like where you are going with this (x0+x1)/2

    Since x1 = x0+dx

    then (x0+x1)/2 = x0+dx/2

    which is the same location as x1-dx/2

  15. #285
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    Red Lines

    Quote Originally Posted by grapes View Post
    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.

    ...
    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 = n3+n*Tn

    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 = n2+Tn

  16. #286
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    Quote Originally Posted by xylophobe View Post
    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.
    Absolutely wrong. It calculates force on a unit test particle, not distance.
    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 = n3+n*Tn
    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.

  17. #287
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    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

  18. #288
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    Jeff, I think that is what he is preparing here:
    Quote Originally Posted by xylophobe View Post
    I like where you are going with this (x0+x1)/2

    Since x1 = x0+dx

    then (x0+x1)/2 = x0+dx/2

    which is the same location as x1-dx/2

  19. #289
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    Calculate

    Quote Originally Posted by grapes View Post
    Absolutely wrong. It calculates force on a unit test particle, not distance.
    ...
    I know that the final result is a force 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:

    Quote Originally Posted by grapes View Post
    I'm happy to do that myself!
    G is the gravitational constant, delta is the density, 2pi y is the circumference of a ring which when multiplied by its cross-section dx dy gives the volume of an infinitesimal ring (multiplied by density gives mass), which is then multiplied by G and divided by distance squared to get the gravitational effect at the apex. However, opposite sides of the ring oppose each other, and cancel some of that force, leaving only the force parallel to the axis, which is found by multiplying by x divided by the distance.

    ...

  20. #290
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    x-dx/2

    Quote Originally Posted by grapes View Post
    Jeff, I think that is what he is preparing here:
    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)

  21. #291
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    Quote Originally Posted by xylophobe View Post
    I know that the final result is a force
    I am aware that you know that, but you still said what you did.
    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:
    The difference is, what I said earlier does not contradict what I said later.
    Quote Originally Posted by xylophobe View Post
    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)
    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.

  22. #292
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    Quote Originally Posted by grapes View Post
    Try using an x-coordinate of x (and a y-coordinate of y)
    to the center of the rectangle.
    YAY!!! Clarity!

    -- 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

  23. #293
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    hmmmmmmm....

    Quote Originally Posted by grapes View Post
    No, multiply that by 6. ...

    Quote Originally Posted by grapes View Post
    ...
    The nice thing about the evaluation of that integral is that the y-integral is a constant! The x variable disappears:



    Multiplying by the other four constants ( ) is still a constant, so the effect of the x-integral is just to multiply that resulting constant by the difference in the x limits. In other words, in your cases, h, each time. So, all four segments have the same answer.
    I multiplied this by "h" as you suggest but it is giving a value that is twice what I expected.

  24. #294
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    Equality

    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.

  25. #295
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    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

  26. #296
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    Quote Originally Posted by xylophobe View Post
    I multiplied this by "h" as you suggest but it is giving a value that is twice what I expected.
    Perhaps I did something wrong. What did you expect, and why did you expect it?
    Quote Originally Posted by xylophobe View Post
    If the double-integral functions as has been described to me then the following integrals should all produce the same answer:
    And they are, it is usually shown early in a calculus class.

    JOOC, though, why did you leave the denominators unscathed?

  27. #297
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    Yes

    Quote Originally Posted by grapes View Post
    ...JOOC, though, why did you leave the denominators unscathed?
    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.

  28. #298
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    If that is giving you the answer you expected, what happened to the "twice"?

  29. #299
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    Mistake

    Quote Originally Posted by grapes View Post
    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.

  30. #300
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    There is no correct way, but some ways are easier than others. My approach from post#194:
    Quote Originally Posted by grapes View Post
    if you're saying that you need that cosine because you make a trig sub to evaluate the integral, be aware that that is the long way! A^2 = y^2 + x^2 is easier, because then A dA = x dx. Actually, I used A = y^2 + x^2
    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.

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