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How to Calculate surface width of a 3d object intersected at an angle?

  1. Jul 28, 2008 #1
    I'm looking for a formula that will give me the surface length of a 3d object when intersected at a specific angle?

    For example if 1" wide object which is 14" long is at 90 degrees (i.e. vertical) to a line, it will be 1" wide when sitting on the line. It will be 14" wide when it is parallel or at 0 degrees to the line. How wide (surface area) will the 1" column be if tilted so it is at 60 degrees to the line?

    Thank you,
    Jim
     
  2. jcsd
  3. Jul 28, 2008 #2

    HallsofIvy

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    Is the "object" you are talking about a rectangle? And when you talk about the width of a '1" column' when it is tilted, are you talking about its projection into the xy-plane? If not, I don't understand what you mean: a '1" column' is 1" wide no matter how it is tilted relative to something else!
     
  4. Jul 28, 2008 #3
    Yes, it is a 3d rectangle, Say, 1" wide x 1" deep x 14" tall. However, I'm only concerned with the width of the rectangle where it meets an intersecting line at an angle.

    If the rectangle was sitting on top of a line at 90 degrees, so perpendicular to the line, like this:

    _|_

    The width of the rectangle where it meets the line would be 1" wide.

    The farther the rectangle is tilted the wider the area becomes where the rectangle intersects the line, like this:

    _/_

    I'm looking for a formula that will tell me if an object is y inches wide and is intersected by a line at x degrees, how wide will the object be at the intersection?

    Some background:
    I'm building a Warren Bridge and the vertical beams are at 60 degrees of the base beam. I know the width of the vertical beams but I need to know how wide they will be when they sit on the base beam at 60 degrees. I know it will be wider than the width of the beam because it is tilted at an angle. I'm guessing about 15% wider, but it would be nice if there were a formula for calculating this?

    Thank you,
    Jim


    _
    _
     
  5. Jul 28, 2008 #4
    That would be the normal width divided by the sine of the angle.
     
  6. Jul 28, 2008 #5
    16* 90/90 = 16 (height)

    16*60/90 = 10.666..... (height)

    you had a rectangle of area 16inches squared
    so now: 16inches squared/10.6666.....inches = 1.5 inches (width)
     
    Last edited: Jul 28, 2008
  7. Jul 29, 2008 #6

    HallsofIvy

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    So you are talking about the projection to the xy- plane. Werg22, the way I am understanding this, it would be the given width times the cosine of the angle: the given width is the hypotenuse of a right triangle and is always larger than the projection.

    taile11, he was asking about width, not area. And your simply multiplying by "angle/90" is not correct.
     
  8. Jul 29, 2008 #7
    *******I changed the heigth from 14" to 16" by accident*******

    HallsofIvy, i uderstand that he is asking about the width. I just found it easy to find it using the area which doesnt change and divide it by the heigth at a particular angle.
    I also realized that i made a big mistake.
    I read your post and found something similar to what you said to be correct. But i figured that instead of multiplying the width by the cosine of the angle, you have to multiply the Heigth by the cosine of the angle.
    why is that?
    well because, like in projectile motion, the 16" remains constant eventhough it changes angles. I consider the 16" to be like the resultant length of it's X and Y components. To more the lesser the angle, the greater the width. Like you had said, the width is represented by the 16" * cosine of the angle. Therefore the width is the X component. X and Y will always change but 16" "resultant" wont.
    so now you have:
    16" * cos(60) = 8" (width)
    16" * sin(60) = 13.854"(height)

    do pythagorean theorem using width and heigth and you still get 16"

    Now, i have this arise question. Do the corners matter? like, it is able to stand on one corner and not be flush to the floor. is yes, the measurements above might not totaly be right.
    so, do they matter?
     
    Last edited: Jul 29, 2008
  9. Jul 29, 2008 #8
    Thank you all for your help. I finally figured it out based on info from this NASA page:

    http://www.grc.nasa.gov/WWW/K-12/airplane/sincos.html

    What I need is a right angle drawn between the two sides of the rectangle, above the line that intersects it.

    Then I can use this formula:
    Cos(angle between a and h) = a/h

    I picked this formula because I know the angle between a and h and I know a.

    I know the angle between a and h is 30 degrees because the support beam (rectangle) is titled at 60 degrees.

    Cos(30) = .866

    .866 = a/h

    .866 = 1/h

    h = 1/.866

    h= 1.1547

    I was thinking it would be around 15% more, but now I know for sure. So the support beam will have a width of just a hair over 15% longer when tilted 60 degrees.
     
  10. Jul 29, 2008 #9
    are you sure?
    as the degrees get smaller, cant you say that the width becomes the height and the height become the width?
    did i misunderstood the problem because that is the way i visualized it(eg: @ angle 0 the width will be 16" and the heigth 1")
    the way that you worked your problem will not conclude with that answer.
     
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