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The length of an uncoiled spring

  1. Nov 8, 2012 #1
    Hi there,

    my question is how would I find the length of an uncoiled spring?

    I hope that is clear enough, but other ways of explaining what i'm looking for look like this: How would I find the length of a line that made a spring x meters high?

    Or how would you know how long a piece of metal must be to manufacture a spring x meters high.

    A little background info. I'm studying architecture and have been thinking about spiral stair cases. Knowing the elevation you hope to achieve via the staircase is useful for knowing how high/how many steps are required. But for knowing how "deep" a step must be (ie the horizontal element that you step on), it seems to me that it would be useful to calculate the entire "length" of your staircase.

    I have considered thinking about the problem as having a step placed at regular intervals on a circle. So you would have x amount of circles stacked on top of each other, with steps that form a spiral. However i'd really like to know if there's some maths that can help me with my problem.

    Thanks in advance for any help you can give! Lntz.
     
  2. jcsd
  3. Nov 8, 2012 #2
    You could try the helix formula and do the arc length of it. Should work.
     
  4. Nov 12, 2012 #3

    HallsofIvy

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    A spring is, as zapz suggested, is a helix. It can be modeled as [itex]x= r cos(t)[/itex], [itex]y= r sin(t)[/itex] [/tex]z= (h/2\pi)t[/tex] where "r" is the radius of the helix and "h" is the vertical distance between two consecutive "turns' of the helix. "t" is the parameter giving each point on the heiix as t varies.

    [itex]dx/dt= -r sin(t)[/itex], [itex]dy/dt= r cos(t)[/itex], and [itex]dz/dt= h/2\pi[/itex].
    The "differential of arclength is
    [tex]\sqrt{(dx/dt)^2+ (dy/dt)^2+ (dz/dt)^2}dt=[/tex][tex]\sqrt{r^2sin^2(t)+ r^2cos^2(t)+ r^2sin^2(t)+ h^2/4\pi^2}dt[/tex][tex]= \sqrt{r^2+ h^2/4\pi^2}dt[/tex]

    The total length is the integral of that from whatever t determines the beginning of the helix to whatever t determines the end of the helix. And, since that differential is a constant, it is just the that constant times the difference in to two "t"s. In particular, if we take z= 0 as the start and z= H as the end, because [itex]z= (h/2\pi)t[/itex] we have [itex](h/2\pi)t= 0[/itex] at one end and [itex](h/2\pi)t= H[/itex] at the other so that t= 0 and [itex]t= 2\pi H/h[/itex] at the other.

    The length of the helix is [itex](2\pi H/h)\sqrt{r^2+ h^2/4\pi^2}dt[/itex]
     
  5. Nov 12, 2012 #4
    Since you are asking about the depth of each step, you find an exact description without considering the length of a helix. Using Halls notation, we have the total height of the staircase = H. Presumably, the height of an individual step is standardized, so lets call it s.

    If N is the total number of steps then N = H/s.

    If you completely neglect the change in height as you go around the staircase, then what happens is that each step takes you a certain angle, A, around the circle. Using Halls' notation, let h be the number of times your staircase wraps around. Then the total angle you go around is 2∏h. Since their are N steps, that means

    2∏h = NA.

    And N = H/s, so A = 2∏hs/H.

    That is the angle of the step. Its horizontal "depth" is equal to the angle times the distance, R, to the central axis of the staircase.

    Depth = R* (2∏hs/H)

    The value of R changes from the inside to the outside of the step. That is why there is more room for you foot on the outer edge of the spiral than on the inner edge.
     
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