Posers and Puzzles

Posers and Puzzles

  1. Subscriberjoe shmo
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    08 Nov '10 03:22
    Originally posted by sonhouse
    Sure, but I am talking about a starting point way up the conic section where it is mostly flat, for instance if you started at the very top of the pool, 7000 mm in diameter and went down 40 mm and used a conic section there, 45 degree angle, the level down there would be 6920 mm. So there would be somewhat less error there than considering a 6000 mm flat ci ...[text shortened]... mm cubed. Where does the difference start? Oh, I misread that, you DID calc a cylinder. Sorry.
    The difference starts when your trying to calculate the rest of the volume excluding the cylinder.

    lets not jump to any conclusion about what happens if we start tweaking parmeters, just try to understand all my calcs using your method...

    my method is a good bit more mathematically involved so we are not going to get into how I got my equation.

    do you understand the calculation of V2?
  2. Subscribersonhouse
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    08 Nov '10 03:261 edit
    Originally posted by joe shmo
    The difference starts when your trying to calculate the rest of the volume excluding the cylinder.

    lets not jump to any conclusion about what happens if we start tweaking parmeters, just try to understand all my calcs using your method...

    my method is a good bit more mathematically involved so we are not going to get into how I got my equation.

    do you understand the calculation of V2?
    Well for starters, is my answer correct for the volume of the conic section as I calculated it in my last post, 15 E6 mm ^3 and change? If not, what is wrong with the way I calculated it? I don't see why it would be different. If I assume the two cylindrical sections and an angle of 45 degrees, it is clear that is a special case where I can simplify the calculation, since it seems to me the 45 degree line cuts the outer cylindrical section exactly in half volume wise. I think, anyway!
  3. Subscriberjoe shmo
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    08 Nov '10 03:301 edit
    Originally posted by sonhouse
    Well for starters, is my answer correct for the volume of the conic section as I calculated it in my last post, 15 E6 mm ^3 and change? If not, what is wrong with the way I calculated it?
    your new method for differences in volumes over 2 is valid for this cone.

    as you can see it is equal to

    (2*pi*R_avg)(W)(h)*(1/2)
  4. Subscribersonhouse
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    08 Nov '10 03:333 edits
    Originally posted by joe shmo
    your new method for differences in volumes over 2 is valid for this cone.
    Yeah, I had to redo my thinking when you mentioned the SLIM possibility I might have been wrong๐Ÿ™‚

    The 45 degree angle thing makes it a special case, easier to calculate.

    So that would mean my error was 100% high. The true value would be 40.53 mm not 41.06.

    I would have seen that eventually๐Ÿ™‚ But I was skyping with my daughter at the same time you and I were both typing tonite so my mind was in two places. In this case, skyping via typing in the comment window, since I have a problem getting my USB mic to connect to the sound card, same with webcam. I'll get that fixed soon though.
  5. Subscriberjoe shmo
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    08 Nov '10 03:38
    Originally posted by sonhouse
    Yeah, I had to redo my thinking when you mentioned the SLIM possibility I might have been wrong๐Ÿ™‚

    The 45 degree angle thing makes it a special case, easier to calculate.

    So that would mean my error was 100% high. The true value would be 40.53 mm not 41.06.
    no the true value is 29.78 mm
  6. Subscribersonhouse
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    08 Nov '10 03:423 edits
    Originally posted by joe shmo
    no the true value is 29.78 mm
    How in the world could the level go from 40 mm to 29 with such a small outer volume?
    Do you mean 39.78?
    At any rate, I think I see my difficulty, you are taking a cylinder, filling it 40 mm deep then expanding from the bottom surface out 40 mm on each side (if you were looking at a horizontal slice) making the level go down by the amount of the volume change from a perfect cylinder. Is that the real story?

    I think maybe we are not talking the same thing. Lets get one thing right: the measured level is 40 mm. Then we want to know the level if the rain used that same volume to fill a straight up cylinder. So if the measurement showed 40 mm exactly, how could less actual rain fill the extra volume?
  7. Subscriberjoe shmo
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    08 Nov '10 03:581 edit
    Originally posted by sonhouse
    How in the world could the level go from 40 mm to 29 with such a small outer volume?
    Do you mean 39.78?
    At any rate, I think I see my difficulty, you are taking a cylinder, filling it 40 mm deep then expanding from the bottom surface out 40 mm on each side (if you were looking at a horizontal slice) making the level go down by the amount of the volume change from a perfect cylinder. Is that the real story?
    I dont know how to make you see this...

    if we had 2 pools , one is just an empty pool (a right circular cylinder of 3.5m radius) the other has our cone section in it.

    in the same rain storm they both collect water over the same area, and they both contain the same volume of water... so there volume per unit area ratios are the same.

    which one has the least height of water?
  8. Subscribersonhouse
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    08 Nov '10 04:085 edits
    Originally posted by joe shmo
    I dont know how to make you see this...

    if we had 2 pools , one is just an empty pool (a right circular cylinder of 3.5m radius) the other has our cone section in it.

    in the same rain storm they both collect water over the same area, and they both contain the same volume of water... so there volume per unit area ratios are the same.

    which one has the least height of water?
    It seems to me they could not have the same volume of water. If you have say 1 mm per minute of rain coming down, in a cylindrical pool after 40 minutes you would have 40 mm of vertical level and a certain volume. If you compare that to a conical pool and it rained 1 mm per minute, after 40 minutes you would have the same volume in both pools? Oh, you said, volume per unit area. Ok, let me gnaw on that one for awhile. Ok, it's been a while. So if a certain unit/min of rain is coming down on a given area, it will have the same volume regardless of what the shape is? So a full cone going down at 45 degrees and 3.5 meters deep like you said, after 40 minutes of rain they would have the same exact volume of rain collected, without spilling over the sides. ok, the same mass of water in both cases. Now what? It seems Aristotle worked this stuff out a couple thousand years ago, right? If the depth of the cylindrical section was also 3.5 meters deep and the volume was the same, the height of water in the cylinder would be 40 mm so how could the depth of water in the cone be less than 40 mm? Wouldn't it be higher than 40 mm? Measuring the depth at the center of the cone of course.
  9. Subscriberjoe shmo
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    08 Nov '10 04:15
    Originally posted by sonhouse
    It seems to me they could not have the same volume of water. If you have say 1 mm per minute of rain coming down, in a cylindrical pool after 40 minutes you would have 40 mm of vertical level and a certain volume. If you compare that to a conical pool and it rained 1 mm per minute, after 40 minutes you would have the same volume in both pools? Oh, you said, volume per unit area. Ok, let me gnaw on that one for awhile.
    They have the same volume of rain. How we measure rainfall is Volume/Area. Since we are assuming the rainfall is uniform, since the areas are the same they are recieving the same volume of rain.
  10. Subscribersonhouse
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    08 Nov '10 04:201 edit
    Originally posted by joe shmo
    They have the same volume of rain. How we measure rainfall is Volume/Area. Since we are assuming the rainfall is uniform, since the areas are the same they are recieving the same volume of rain.
    So the cylinder section would have a lower depth than the cone. And that would be the same for a conical section Vs a cylinder also. Makes sense now.
  11. Subscriberjoe shmo
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    08 Nov '10 04:22
    Originally posted by joe shmo
    They have the same volume of rain. How we measure rainfall is Volume/Area. Since we are assuming the rainfall is uniform, since the areas are the same they are recieving the same volume of rain.
    Now since we know they get the same volume of rain, but we know that the true height of rain only comes from the shape with constant crassectional area in the direction of height (ie the right circular cylinder) we say

    Vcone= Vcylinder

    1.4615*10^9 mm^3 = pi*(3500mm)^2* h

    h=(1.4615*10^9*mm^3)/(pi*3500^2*mm^2)
  12. Subscribersonhouse
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    08 Nov '10 04:313 edits
    Originally posted by joe shmo
    Now since we know they get the same volume of rain, but we know that the true height of rain only comes from the shape with constant crassectional area in the direction of height (ie the right circular cylinder) we say

    Vcone= Vcylinder

    1.4615*10^9 mm^3 = pi*(3500mm)^2* h

    h=(1.4615*10^9*mm^3)/(pi*3500^2*mm^2)
    Where do you get that 1.46E9 # ? I get 1.53E9 for a cylindrical volume 40 mm high.
    I just calc'd it exactly as you wrote it, PIX 3500X3500X40.

    At any rate, thanks for sticking with me on this.
  13. Subscriberjoe shmo
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    08 Nov '10 04:392 edits
    Originally posted by sonhouse
    Where do you get that 1.46E9 # ? I get 1.53E9 for a cylindrical volume 40 mm high.
    I just calc'd it exactly as you wrote it, PIX 3500X3500X40.

    At any rate, thanks for sticking with me on this.
    V1=pi*(Rmin)^2*h_cone=pi*(3000 mm)^2 * (40 mm) = 1.13097*10^9 mm^3

    the volume of the remaining part of the cone is:

    V2= 2*pi*R_avg *W*H*1/2 = pi*3020mm*40mm*40mm=1.51802*10^7 mm^3

    V_total=V1+V2 = 1.14615*10^9 mm^3

    there was a typo in the other post

    And your welcome ๐Ÿ™‚

    thats all for tonight for me however, so good night.
  14. Subscribersonhouse
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    08 Nov '10 04:57
    Originally posted by joe shmo
    V1=pi*(Rmin)^2*h_cone=pi*(3000 mm)^2 * (40 mm) = 1.13097*10^9 mm^3

    the volume of the remaining part of the cone is:

    V2= 2*pi*R_avg *W*H*1/2 = pi*3020mm*40mm*40mm=1.51802*10^7 mm^3

    V_total=V1+V2 = 1.14615*10^9 mm^3

    there was a typo in the other post

    And your welcome ๐Ÿ™‚

    thats all for tonight for me however, so good night.
    for sure, we can do this again tomorrow!
  15. SubscriberAThousandYoung
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    09 Nov '10 07:291 edit
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