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Area of trapezoid formed by slicing a cylinder

  1. May 29, 2015 #1
    What is the equation of the area of the trapezoid when a cylinder of radius R is cut by a plane inclined at an angle α? (Orange area of the figure)

    How can I relate it with Abase=π*R2 or with Ainside=2*R*L?

    Any hints please?

    Thank you very much in advance.
    cylinder.png
     
  2. jcsd
  3. May 29, 2015 #2

    berkeman

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    That looks pretty straightforward unless I'm missing something. Have you taken trig yet? :smile:
     
  4. May 29, 2015 #3
    Yes, I think I have but, unless I'm not grasping the whole problem, it's not quite simple.

    I know that one of the sides of the trapezoid is 2R, but I don't know anything else.
    I first thought that the length of the trapezoid would be L*cosα, but I think taht L*cosα it's only the projection of the length of the trapezoid in the xy plane (the plane of the screen of the computer).
     
  5. May 29, 2015 #4

    Mark44

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    Draw a line from the center of the left end of the cylinder to the center of the chord on the right end of the cylinder. Using the angle ##\alpha## you should be able to use trig to find the vertical distance (h in the figure below) between the central axis of the cylinder to the center of the chord.

    Once you know h, you can find the coordinates of the chord on the circular end of the cylinder.
    Snapshot.jpg
     
  6. May 29, 2015 #5

    Ibix

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    I don't think the orange region is a trapezoid. If you imagine extending the cylinder in both directions, the orange slice extends into an ellipse (get a sausage, cut it on the diagonal and look at the face you expose). Reducing the cylinder back to its original length simply trims the ends off the ellipse. So the ends of the shape are straight lines, but the sides are curves. Integration may be needed.
     
  7. May 29, 2015 #6

    SteamKing

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    It's not clear what you are calling Abase.

    In any event, if you constructed a true size view looking directly at the plane cutting thru the horizontal cylinder, you would see that the long sides of the "trapezoid" were not straight lines; in fact, these sides should be curved, possibly parts of a parabola.

    One end of the "trapezoid" has length 2R only because of its vertical position on the end of the cylinder. At the opposite end, the length of the "trapezoid" is less than 2R.

    If you make a series of cuts along the length of the cylinder, you can calculate how wide the "trapezoid" is at any arbitrary point. If you have the width at the ends and in the middle of the cylinder, you can use Simpson's Rule to calculate the area pretty accurately.
     
  8. May 29, 2015 #7

    berkeman

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    Can you give us some context to this question? Is it for a real-world application, or just a curiosity? Or is it for schoolwork? Thanks.
     
  9. May 29, 2015 #8

    RUber

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    I would integrate with respect to the height of the "trapezoid," i.e. ##A = 2\int_0^H w(x) dx ## where H is the total height of the shape and w is the half-width of the "trapezoid".

    If you can get it into the form of
    ## A = m\int_0^H \sqrt{c^2 - x^2} dx ##
    then the integral evaluates to:
    ##\frac m 2 \left( H \sqrt {c^2 -H^2}+c^2 \sin^{-1} ( H/c ) \right) ##
    I had to look that one up, so maybe it'll save you some time.
     
  10. May 29, 2015 #9

    Mark44

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    My interpretation of this problem, which could be wrong, is that the diameter at the left end of the cylinder and the chord at the right end, define a trapezoid. In my interpretation, it's immaterial whether the nonparallel sides of the trapezoid are inside the cylinder or outside it.

    The problem is somewhat ambiguous. The OP originally had "Area of ellipse..." in the title, but in post 1, asked about the area of the trapezoid. I changed the thread title to be consistent with the question in post 1.
     
  11. Jun 1, 2015 #10
    Yes, of course. It's a model for a real-world application.

    Imagine a magnetic flux (or any type of flux, for that matter) that passes through that cylinder.
    If the cylinder's axis is parallel to the x axis and parallel to the magnetic flux vector, the area crossed by the flux is the Abase=π*R2, and the length of the path of the magnetic flux is L. On the other hand, if the magnetic flux vector is orthogonal to the cylinder, the area crossed by the flux is Ainside=2*R*L and the length of the path of the magnetic flux is 2*R, it's diameter.

    Now, when the cylinder is inclined with an angle α to the x axis, the magnetic flux through the cylinder must be decomposed in two parts:

    -one is the projection of the cylinder in the x axis, where the area crossed by the flux is the area of an ellipse (Abase/cos(α) and the length of the path of the magnetic flux is L*cos(α);
    -the other is the projection of the cylinder in the y axis (orthogonal) where the area crossed by the flux is in my opinion the area of the trapezoid that I want to find (and that I need/I think to have a sin(α) in the expression) and the length of the path of the magnetic flux is L*sin(α);

    Thanks for your help.
     
  12. Jun 1, 2015 #11
    It was my mystake, the new titles fits much better. Thanks!
     
  13. Jun 1, 2015 #12
    Abase is the area of the base of the cylinder, hence π*R2.

    Your chain of thought is similar to mine, but I've lost myself along your description. Could you be so kind to give me some start on the expressions?
    As I said, I think that the area of the trapezoid is somehow related (or I wish it'd be for a matter of easyness of the treatment of the expressions) with the sin(α).
     
  14. Jun 1, 2015 #13
    Thanks for your help.
    I'm sorry but I don't understand. When you consider w the half-width of the trapezoid, are you refering to the length of the base of the trapezoid? In this case which base? The minor or the major?
     
  15. Jun 1, 2015 #14
    It's actually the projection of the inclined cylinder on the y-z plane. (the side view)
    The projection is in the x-y plane (the top view). What you're doing is cutting a straight cylinder with an inclined plane and taking the projection of the cut-section. As I understand it, what you want is the projection of a cylinder that is inclined. The two are very different things! Think about viewing a cylinder that is inclined with the x-axis from the top versus what you were doing. The top view of a tilted cylinder is simply a rectangle with sides ##2R## and ##Lcos(\alpha)##.
     
    Last edited: Jun 1, 2015
  16. Jun 1, 2015 #15

    RUber

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    In my breakdown of it, I was calling w the half-width of the short side of the trapezoid.
    upload_2015-6-1_9-31-14.png
     
  17. Jun 1, 2015 #16
    I see and agree.
    In that case my only last question is what is c.
     
  18. Jun 1, 2015 #17
    Would both of you care to explain, exactly how is the section of the cylinder relevant when you're actually tilting the cylinder itself ?
     
  19. Jun 1, 2015 #18
    Imagine slicing the tilted cylinder from one base to another. That's how the flux "see/crosses" the cylinder when it is tilted.
     
  20. Jun 1, 2015 #19
    ?? As lbix pointed out. The plane figure is not a trapezoid but a truncated ellipse. ??

    ---Unless it's the trapezoid contained within the truncated ellipse that is of interest.
     
    Last edited: Jun 1, 2015
  21. Jun 1, 2015 #20

    RUber

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    c is a constant for the formula, just like m. When I worked it out, it ended up being some combination of initial parameters R and sin(alpha).
     
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