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

  1. #241
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    Grapes: looks like we both fumbled that potential function.

    Xylo: be advised that delta(x) refers to a certain number of slices. Strictly speaking, dx is an infinitesimal. There is not a certain number of dx slices. Dx is used in integration while delta(x) is for numerical approaches. They're not handled the same way.

    That quote even says that we sum an infinite number of dx slices to get the integral. Your method can never get to infinite slices. You'll run out of computer processing power and memory trying to get an exact answer.

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    Last edited by publiusr; 2012-May-14 at 10:15 PM.

  3. #243
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    Infinitesimals Defined

    Quote Originally Posted by ShinAce View Post
    ...

    Xylo: be advised that delta(x) refers to a certain number of slices. Strictly speaking, dx is an infinitesimal. There is not a certain number of dx slices. Dx is used in integration while delta(x) is for numerical approaches. They're not handled the same way.

    That quote even says that we sum an infinite number of dx slices to get the integral. Your method can never get to infinite slices. You'll run out of computer processing power and memory trying to get an exact answer.
    "In common speech, an infinitesimal object is an object which is smaller than any feasible measurement, but not zero in size; or, so small that it cannot be distinguished from zero by any available means. Hence, when used as an adjective, "infinitesimal" in the vernacular means "extremely small"."

    Please notice that it states that the infinitesimal is "not zero in size".

    Because "dx" is not a zero value then dx/2 (or any fraction of "dx") is not a zero-valued amount/size.

    Leibniz notation uses the fraction "dy/dx" and it should be noted that "dx" is multiplied by the function and if the function is just a constant then 2*dx is acceptable as well as 1/2*dx.

  4. #244
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    Quote Originally Posted by xylophobe View Post
    I do not want readers to get the wrong perception (whether by me or someone else) of how integration works so I will direct to:

    "By the logic above, a change in x (or Δx) is the sum of the infinitesimal changes dx. It is also equal to the sum of the infinitesimal products of the derivative and time. This infinite summation is integration; ..."

    The above text and other readings on the Fundamental theorem of calculus seem to agree with my description of how the value of "x" advances by increments of "dx".

    Method of exhaustion.
    You missed indicating that the quote is from a paragraph on Physical Intuition rather than one of formal definitions.
    You're describing an analogy not the thing in itself.

    In general it looks like you haven't really understood how the theory of limits is applied to get from sums of slices to integrals, which definitely aren't sums of slices..

    "In common speech" is a large red flag being waved frantically to show that this is NOT mathematics.

    It approaches mathematics as it vaguely alludes to how limits are defined, but without the formal definition, it's nearly useless to get any insight from.

    The mathematical version doesn't talk about measurements and quite easily distinguishes from zero.
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  5. #245
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    What Henrik said!

    From that same wiki page:
    "The method of indivisibles related to geometrical figures as being composed of entities of codimension 1. John Wallis's infinitesimals differed from indivisibles in that he would decompose geometrical figures into infinitely thin building blocks of the same dimension as the figure, preparing the ground for general methods of the integral calculus. He exploited an infinitesimal denoted 1/infinity in area calculations."

    You're plugging a number into dx, namely, 1. Try plugging 1/infinity and see what you get. You get 0. That's why the method of infinitesimals applies to integral calculus, not indivisibles related to geometrical figures.

    Here's an example.

    0.1111111111111periodical is 1/9.
    Take that, multiply by 9.
    You now have 0.99999999999999999periodical.
    But 1/9 times 9 is 9/9. Which is equal to 1.
    What's the difference between 0.9999999periodical and 1? We can label it as dx and call it an infinitesimal. Under no circumstances am I allowed to plug any number other than 0 in place of dx. We keep saying it's 0 because you try to make it 1. We're not defending dx as nothing, we're opposing the fact that you continuously replace it with an integer and try to say that this applies to integral calculus. It does not apply to integrals.

    As it's own thread, the difference between 9/9 and 1 drew much debate. But the point remains, the difference between 0.9999999... and 1 is determined by how far down the periodical you're willing to go. Go forever, and the difference is an infinitesimal. There's no point trying to pin a number on the difference, it's a concept. Use basic arithmetic and you find that 1 - 9/9 = 0 .

    Now we're getting sidetracked again.

    Also, make note that everyone that has done integral calculus has used dx/2 at some point. It shows up in a change of variable or even integration by parts.
    Let x = 2u. Then dx = 2du , or du = dx/2 .
    When you do that, you get rid of dx, and work in du. When you're done, you back substitute 2u for x to get the answer in terms of x. This is something that you would do to make an integral 'doable'. You would use it to reduce the expression into something simpler.

    You're not doing a change of variable and getting rid of the original variable. You're trying to redefine an existing variable. X does not equal 2X.
    You're not using infinitesimals. You're using plain old differences. Aka, deltas. Writing dx to mean delta(x) is only going to confuse the reader.

    Imagine I want to build a staircase. It needs to be 10 feet deep and 10 feet high. Since all edges are right angles, the total length of wood to make the staircase is 20 feet(we're obviously ignoring the width of the staircase). Whether there is 1, 10, 100, or 1000 steps, the total length of wood to make the staircase is 20 feet. But, in calculus, if I let the number of steps be infinity, the right angles disappear and the total length of wood needed is 14.14 feet. That's because it's not even a staircase anymore. It has become a flat ramp. Is your cone like the staircase or the ramp?

    Hmmm... even if I had a trillion steps, I would need 20 feet of wood. But if each step was an infinitesimal, I could build the staircase using under 15 feet of wood. It would appear that the two ideas cannot be treated the same way. Yet, by your logic, if the wood I need does not change despite the number of steps, then infinite steps will need 20 feet of wood. This abuse of limits proves nothing.

  6. #246
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    Quote Originally Posted by ShinAce View Post
    Grapes: looks like we both fumbled that potential function.
    No, we didn't.
    Quote Originally Posted by ShinAce View Post
    Hmmm... even if I had a trillion steps, I would need 20 feet of wood. But if each step was an infinitesimal, I could build the staircase using under 15 feet of wood. It would appear that the two ideas cannot be treated the same way. Yet, by your logic, if the wood I need does not change despite the number of steps, then infinite steps will need 20 feet of wood. This abuse of limits proves nothing.
    It's still 20. You have one chance to redeem yourself.

    Analyze his diagrams and show that his error does not increase.

  7. #247
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    Two off-topic questions

    What should the "d" in "dx" be thought to stand for?

    When building a staircase of infinitesimally-thin wood
    (like paper! It can fold!), ten feet high and ten feet deep,
    why does the length of wood (or paper) needed decrease
    from 20 feet to ten times the square root of 2 feet when
    the number of steps becomes infinite?

    -- Jeff, in Minneapolis
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    "I find astronomy very interesting, but I wouldn't if I thought we
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  8. #248
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    Oh, wow! That was a mistake by ShinAce? I wondered
    how the length could change like that.

    -- 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

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    Because infinite steps does not make any sense. If you can see that there are steps, there aren't an infinite number of them. If there's an infinite number of them, you can't see if they are steps or slopes. So what does infinite steps imply? If you imagine a bunch of little steps, you say 20 feet of material are required. I ignore that there are any steps at all and just make a ramp. You would need 20 feet because you want to build a staircase with n steps. I would need 14.14 feet because I'm not building steps, I'm making a ramp. I just lay one piece of wood down and call it a day. Don't take the analogy too seriously. I'm not saying that numerical integration gives an answer different than integral calculus. If you really are making little steps, then you need 20 feet of material. If you think infinite steps requires 20 feet, go ahead and buy 20 feet of wood. Your staircase will look and feel the same as my ramp, and I'll save some money on material.

    Grapes: the mistake I made was assuming that the center of gravity coincides with the gradient of the potential function being zero. Your mistake was when you said that the potential of a sphere is lowest at the surface. That's true of a spherical shell of mass. But here, the cone is not a shell, it is a solid. Gravitational potential of a solid sphere is lowest at its center. Whether or not you're allowed to go through the sphere's surface is still arbitrary.

    Be advised I briefly analyzed the error way back in post 209. I'm not going to redo the work.

    dx can be read as "an infinitesimal length along x". It has the same units as the x axis.

  10. #250
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    Quote Originally Posted by ShinAce View Post
    Because infinite steps does not make any sense. If you can see that there are steps, there aren't an infinite number of them. If there's an infinite number of them, you can't see if they are steps or slopes. So what does infinite steps imply? If you imagine a bunch of little steps, you say 20 feet of material are required. I ignore that there are any steps at all and just make a ramp. You would need 20 feet because you want to build a staircase with n steps. I would need 14.14 feet because I'm not building steps, I'm making a ramp. I just lay one piece of wood down and call it a day. Don't take the analogy too seriously. I'm not saying that numerical integration gives an answer different than integral calculus. If you really are making little steps, then you need 20 feet of material. If you think infinite steps requires 20 feet, go ahead and buy 20 feet of wood. Your staircase will look and feel the same as my ramp, and I'll save some money on material.
    , with infinitesimal steps you still need 20 feet of wood.

    This time you're the one using intuition rather than formalism to argue your case.
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  11. #251
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    dx

    If dx=0 then any number "n" multiplied by "dx" will equal zero.

    0=dx*2
    0=dx*5
    0=dx*500
    0=dx*googolplex
    0=dx*n
    1=dx*infinity

    If dx=0 then the limit of "dx*n" as "n" approaches infinity somehow changes from zero to some non-zero value (I'll just say the number "1" for discussion purposes). The limit from both directions always equals zero except when n=infinity which is when dx=1 This seems to be the basis of the dx=0 argument.

    Intuitively, the theorem simply states that the sum of infinitesimal changes in a quantity over time (or over some other quantity) adds up to the net change

    For calculus a definite integral is defined over an interval [a,b] with a lower bound "a" and an upper bound "b" which is supposedly divided by the infinitesimal called "dx" so therefore there must be a non-zero value for "dx" or else the integral will always stay at the lower limit and "x" will never advance to the upper bound.

    The use of infinitesimals in Leibniz relied upon a heuristic principle called the Law of Continuity: what succeeds for the finite numbers succeeds also for the infinite numbers and vice versa.

    Despite these attempts calculus continued to be developed using non-rigorous methods until around 1830 when Augustin Cauchy, and later Bernhard Riemann and Karl Weierstrass, redefined the derivative and integral using a rigorous definition of the concept of limit.

    Also, read the geometric intuition: "...with this approximation becoming an equality as h approaches 0 in the limit."

  12. #252
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    Again, don't take the analogy too seriously. Formal definitions of mathematics have failed over and over in this thread. Why beat a dead horse?

    Functions rarely look like staircases. They look like ramps. In my example, what we're doing is finding the arc length of the hypotenuse. That arc length is the wood that you need. The point I'm trying to make is that if you picture it like a staircase that just gets finer and finer, you get those nonsense answers. It actually starts as a ramp(the slope of the cone), and then gets approximated as a staircase. When we do integral calculus, we work out the area between the ramp and the floor below it. I take the height of the ramp at a point along the floor below it. That's my function. Multiply this by the width of that point, dx, and integrate from where the ramp meets the floor to where the ramp meets the next level floor.

    Luckily, it's easy to see that it's a triangular prism and the volume is simply the width of the staircase, times its length, times the height, divided by 2.

    When we setup an integral, we rarely work with stepped functions. Just like the cone, they are continuous functions. The width of each slice(the width of each stair) is simply dx. When a first year calc teacher draws it on the blackboard, they draw what looks like a staircase. All that is supposed to show is that a staircase is a decent approximation. We still need a better one. Luckily, infinitesimal steps gives us the exact answer we're looking for.

    If you can show that the sloped side of the cone has a length better approximated by the length of wood in a fine 'staircase', please do so. You've shown that the limit of 20 is 20, which is nice. How about for for arc length? For me, I use arc length, dx, and integral calculus. I don't substitute dx for integers. I don't add extra terms because the function isn't a flat line. If I want to do a sum, I'll do a Riemann sum and use the midpoint rule. I don't need to break the cone into a staircase, solve that, and then add a bunch of little triangles so that my staircase looks like a ramp again.

    At any rate, we're still talking about dx when it's irrelevant. If you want to use dx to mean xn-xn-1 and work out the numerical approximation, be my guest. I'm not doing it. There's so much work that has gone into this thread for so little gain.

    You want formal? Deal! Time to sidetrack this thread beyond recognition.

    I have a function, y=x. It's a straight line, so it's continuous. The area under the curve from x=0 to x=5 is 25/2.
    Now we add the exception that at the point x=3, y=0. It's no longer continuous. What's the area under the curve from x=0 to x=5?
    What's that? 25/2? Correct! What does your intuition say? That there's a hole in the function(it drops to zero there) so the area must be lower. What if instead of y=0, y=infinity at x=3. It's still 25/2? Wow! That's crazy talk!

    Formally, integrals don't care about points. The integral of a given function from [0,5] = [0,5) = (0,5] = (0,5).

    What is the thickness of a point anyway? Is it zero? Is it infinitesimal in size? Is it 1?

  13. #253
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    Quote Originally Posted by ShinAce View Post
    ... Multiply this by the width of that point, dx, and integrate from where the ramp meets the floor to where the ramp meets the next level floor. ...
    The sum of all the "dx" segments should equal the integral's upper limit minus the lower limit.

    Maybe instead of stating that

    1=dx*infinity

    I should have written

    (Upper Limit - Lower Limit) = dx*infinity

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    Why should it even matter? That's what the fundamental theorem of calculus is for. Integrate, then evaluate at the upper limit, substract the evaluation at the lower limit, and voila, the answer.

    The sum of all the "dx" segments should equal the integral's upper limit minus the lower limit.
    ...if and only if the number of segments is infinite.

    Who cares what dx*infinity should be? If you define dx as 1/infinity, then dx*infinity gives you 1 along the x axis. You could define it as 5/infinity. Now what? Do we scale the limits of integration into dx? Or is it the entire x-axis/infinity? So that dx*infinity=x-axis?

    I'll say it again. Dx is an infinitesimal. Take the height of the cone, divide by infinity, and you get infinitesimal disks. h/infinity = dh
    If you divide the height into n slices, you have h/n = h/n. If you want dh, then you have to make n=infinity.

    Why waste your time arguing about infinitesimals when you're trying to accomplish a numerical integration?

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    Significance of the Error

    Quote Originally Posted by ShinAce View Post
    ...Why waste your time arguing about infinitesimals when you're trying to accomplish a numerical integration?
    This part of the discussion is important because if dx=0 then the location of the CoG for the rings is meaningless because dx/2=0 also.

    I think the unique location of the CoG for each of the rings makes a difference in the outcome of the calculation for gravitational influence upon the point-mass "P".

    If dx/2 is a non-zero value then due to the greater-than-the-square increase in the number of rings the disappearance of the error does not happen.

    Previously it was stated that the error dx/2 gets smaller and smaller as the number of divisions of the integral interval [a,b] increases and by this reasoning the result of the double-integral approaches an exact and correct value but the analysis in this thread shows that even though the size of "dx" might decrease to half it size (and the error decreases to half its value) the quantity of errors increases by 22+T2 (for segment 2, the first frustum)

    The size of the error decreases directly proportional to the number "n" of equal-sized divisions of [a,b] while the quantity of errors increases by a "greater-than-the-square" rate.

    Segment 2 Number of Rings/Errors = n2+ Tn

  16. #256
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    The conflict there is between mathematics and numerical analysis.

    When doing the calculations with a computer, once the slices are small enough, the errors of the computations does indeed start to accumulate faster than the error of the approximation drops, so the limit of the numerical approximation is not the limit of the ideal mathematical solution.
    This is basic to numerical analysis, you can only ever expect to get an approximate answer, but if you do the analysis of the algorithm right you'll also know the error bars of your result.
    And it allows you to do problems that can't be handled trivially by "real" mathematics as the cone can.
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  17. #257
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    Quote Originally Posted by xylophobe View Post
    If dx/2 is a non-zero value then due to the greater-than-the-
    square increase in the number of rings the disappearance of
    the error does not happen.
    This statement shows that your analysis is fundamentally
    wrong. You basically have no idea what you are doing.
    You need to form a completely new understanding.

    Unfortunately I'm not knowledgeable enough to guide you
    toward a correct understanding. ShinAce and grapes are
    knowlegeable enough, but so far they seem either to not
    realize the error you are making, or haven't explained it
    clearly enough for you and I to understand.

    Quote Originally Posted by xylophobe View Post
    Previously it was stated that the error dx/2 gets smaller and
    smaller as the number of divisions of the integral interval [a,b]
    increases and by this reasoning the result of the double-integral
    approaches an exact and correct value but the analysis in this
    thread shows that even though the size of "dx" might decrease
    to half it size (and the error decreases to half its value) the
    quantity of errors increases by 22+T2
    (for segment 2, the first frustum)
    Your mistake is revealed by the words "quantity of errors".

    I could try to explain what is wrong with that notion, but I
    hope that both ShinAce and grapes will do so, because I do
    not know how.

    -- 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

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    It's already been done, Jeff.

    You'd have to go back to posts 201 and 209. In 201, xylophobe posts a nice drawing. I say nice because we don't even need numbers to talk about it. It shows a frustrum cut into 1 slice, then 2 slices, then 4 slices.

    As a single slice, there is a red triangle showing the error portion. As two slices, we see two red triangles and a red rectangle. The rectangle isn't part of the error, just the two triangles. Then with the frustrum in 4 pieces, the error is the 4 smaller triangles drawn.

    Print it on to a transparency, overlay them, and tell me the error from 4 slices is bigger than 1 slice. I'll roll my eyes so quickly you'll think I'm having a seizure. I might actually go into a seizure.

    Just look at it! You can clearly see that as you cut the cone into more slices, the triangles get smaller. Plus it's obvious that the new triangles are always within the original large triangle(1 slice case) while they cover a smaller area. In mathematical language, as the number of slices increases, the new error is a subset of the old error. Also, the old error is not a subset of the new error. This says that the error gets smaller as we use more slices. The number of error terms is the same as the number of slices.

    The size of the error decreases directly proportional to the number "n" of equal-sized divisions of [a,b] while the quantity of errors increases by a "greater-than-the-square" rate.
    Getting into the actual math, and laying out the error as a monotonic decreasing sequence isn't going to help. Forget about math for a second. Why are there any red rectangles in the error at all? There aren't! the error is always composed of those red triangles.

    When you make an error in analyzing error, you get a mistake, not increasing error.

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    Two Errors

    Quote Originally Posted by ShinAce View Post
    It's already been done, Jeff.

    You'd have to go back to posts 201 and 209. In 201, xylophobe posts a nice drawing. I say nice because we don't even need numbers to talk about it. It shows a frustrum cut into 1 slice, then 2 slices, then 4 slices.

    As a single slice, there is a red triangle showing the error portion. As two slices, we see two red triangles and a red rectangle. The rectangle isn't part of the error, just the two triangles. Then with the frustrum in 4 pieces, the error is the 4 smaller triangles drawn.

    Print it on to a transparency, overlay them, and tell me the error from 4 slices is bigger than 1 slice. I'll roll my eyes so quickly you'll think I'm having a seizure. I might actually go into a seizure.

    Just look at it! You can clearly see that as you cut the cone into more slices, the triangles get smaller. Plus it's obvious that the new triangles are always within the original large triangle(1 slice case) while they cover a smaller area. In mathematical language, as the number of slices increases, the new error is a subset of the old error. Also, the old error is not a subset of the new error. This says that the error gets smaller as we use more slices. The number of error terms is the same as the number of slices.



    Getting into the actual math, and laying out the error as a monotonic decreasing sequence isn't going to help. Forget about math for a second. Why are there any red rectangles in the error at all? There aren't! the error is always composed of those red triangles.

    When you make an error in analyzing error, you get a mistake, not increasing error.
    Your last post shows that you do not understand the problem.

    There are two errors that have been discussed: one involving the mass calculation and one involving the CoG calculation.

    I have already explained that the mass error is a decreasing error and that it vanishes as the size of "dx" approaches zero. That is what the red triangles and rectangles show. It proves that the mass calculation is accurate.

    The second error is in the location of the CoG for the rings. This error does not vanish and is the reason the double-integral does not deliver a correct solution.

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    Telling me I don't understand is plain rude. For that reason, I won't be offering you any help.

    Symmetry clearly dictates that the center of mass of a ring is along the central axis. That means...

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    He was probably more sure that he was right when he
    told you that you don't understand than I was sure I was
    right when I told him he doesn't understand.

    I'd like to see an explanation of what is wrong with his
    thinking that he and I can both understand.

    I don't understand why he thinks the location of the
    center of gravity of the rings has a separate error.

    I don't understand why he thinks that error increases
    just because the number of rings increases.

    I don't understand whether you have addressed those
    two specific questions.

    -- 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

  22. #262
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    Jeff, you obviously see the hole. The "cog calculation" is not a cog calculation. It's several pages back but as you know, the center of gravity is found when applying a force at a point exerts no torque. Torque. Such a beautiful word, all french looking and all.

  23. #263
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    Error: Location of CoG for Each Ring

    Quote Originally Posted by Jeff Root View Post
    He was probably more sure that he was right when he
    told you that you don't understand than I was sure I was
    right when I told him he doesn't understand.

    I'd like to see an explanation of what is wrong with his
    thinking that he and I can both understand.

    I don't understand why he thinks the location of the
    center of gravity of the rings has a separate error.

    I don't understand why he thinks that error increases
    just because the number of rings increases.

    I don't understand whether you have addressed those
    two specific questions.

    -- Jeff, in Minneapolis
    In the below picture I show the same frustum being divided by 1, then 2, then 4 or in the language we have been using dx=h, then dx=h/2, then dx=h/4

    The quantity of rings (and errors) increase as I have already posted:

    1 slice ==> 2 rings = (1+1) = 12+T1
    2 slices ==> 7 rings = (2+1)+(2+2)= 22+T2
    4 slices ==> 26 rings = (4+1)+(4+2)+(4+3)+(4+4)= 42+T4
    8 slices ==> 100 rings = (8+1)+(8+2)+(8+3)+(8+4)+(8+5)+(8+6)+(8+7)+(8+8) = 82+T8
    n slices ==> n2+Tn

    Tn = Triangular number (the sum of numbers 1+2+3+4+...+n)

    The error in the location for the CoG is shown by the heavy red lines and one error is present for each ring. This error has a component along the x-axis and a component along the y-axis. For simplicity's sake I have only been discussing the x-axis component of these errors.


    Click image for larger version. 

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  24. #264
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    xylophobe,

    I'll start with a question that I have been waiting and
    waiting and waiting for you to explain:

    What does the upper-right corner of those rings have to
    do with anything?

    -- 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

  25. #265
    Join Date
    Dec 2004
    Posts
    12,587
    xylophobe,

    I modified your last illustration. The way you are
    trying to analyze it, the error for each iteration is
    proportional to the sum of (the length of each red
    line times the area of the pink rectangle that the
    red line is in).

    -- Jeff, in Minneapolis
    Attached Thumbnails Attached Thumbnails Click image for larger version. 

Name:	frustana.png 
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ID:	16858  
    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. #266
    Join Date
    May 2005
    Posts
    5,720
    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.

  27. #267
    Join Date
    Jul 2006
    Posts
    221

    Upper Right Corner

    Quote Originally Posted by Jeff Root View Post
    xylophobe,

    I'll start with a question that I have been waiting and
    waiting and waiting for you to explain:

    What does the upper-right corner of those rings have to
    do with anything?

    -- Jeff, in Minneapolis
    The upper right corner of the rectangles (or rings) is the location used by the double-integral for the calculation of the distance and also for the calculation of the cosine (which uses "x" divided by the distance). To be correct "x" should be the location of the CoG for the rings and the distance should be from point-mass "P" to the CoG for each ring.

    As I discussed in post 189 the location of the CoG for each rectangular ring (dx*dy) is in the center of each ring (I am ignoring the triangular rings for simplicity). I also show in post 189 the correct formula to get the correct distance from point-mass "P" to the CoGs. The correct "x" location for the cosine calculation is just x-dx/2

    Just as "y" is related to "x" by the formula y=xr/h so too is "dy" related to "dx" by dy=dxr/h

    Click image for larger version. 

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    The dissection of the grapes double-integral into its component parts:



    Constants:

    Distance:

    Inverse Square of Distance:

    Cosine:

    Volume of a Cone:

  28. #268
    Join Date
    Jul 2006
    Posts
    221

    Sum of the Cubes

    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.
    The sum of the outer edges of the larger cube (now composed of 8 smaller cubes) remains the same = 12.

    The sum of all the edges increases because you essentially add interior edges to the large cube.

  29. #269
    Join Date
    May 2005
    Posts
    5,720
    Quote Originally Posted by xylophobe View Post
    The upper right corner of the rectangles (or rings) is the location used by the double-integral for the calculation of the distance and also for the calculation of the cosine (which uses "x" divided by the distance). To be correct "x" should be the location of the CoG for the rings and the distance should be from point-mass "P" to the CoG for each ring.
    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.
    Quote Originally Posted by xylophobe View Post
    The sum of the outer edges of the larger cube (now composed of 8 smaller cubes) remains the same = 12.

    The sum of all the edges increases because you essentially add interior edges to the large cube.
    I was hoping that was obvious!

  30. #270
    Join Date
    Dec 2004
    Posts
    12,587
    Quote Originally Posted by xylophobe View Post
    The upper right corner of the rectangles (or rings) is the
    location used by the double-integral for the calculation
    of the distance and also for the calculation of the cosine
    (which uses "x" divided by the distance). To be correct
    "x" should be the location of the CoG for the rings and
    the distance should be from point-mass "P" to the CoG
    for each ring.
    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
    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

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