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Aerospace Ionocraft shape

  1. Jun 14, 2016 #1


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    Gold Member

    Hmmm maybe this goes here, maybe it doesn't.
    Here is a picture that depicts basically what I'm talking about.

    From what I understand, Ionocrafts work on N3L, and shoot ions down, pushing the craft up. This makes me think that the more charges you have, the more thrust you can get. This makes me think you want lots of charges and low weight. (stop me when I mess up)
    This makes me think for the shape of the base, you would want a high cross-sectional area to perimeter ratio. I keep seeing things about equilateral triangles being used.
    ##\frac{\pi r^2}{2 \pi r} =\frac{r}{2}##
    Square (half the length of a side is r):
    ##\frac{(2r)^2}{4(2r)} =\frac{r}{2}##
    equilateral triangle (height is r, b is 1/2 side):
    ##\frac{2*1/2*b*r}{3*2*b} = \frac{r\frac{r}{\sqrt{3}}}{3*\frac{2r}{\sqrt{3}}} = \frac{r}{6}##

    So, hold the distance from the wire on top to the plane constant, and approximate the field inside the ionocraft as constant (or at least the vertical component, the outwards components will cancel because symmetry). You have equal charges/unit mass (or per unit length) in a circle and square. However, you have 1/3 as many charges with the equilateral triangle (per unit length of material i.e. per unit mass), and thus 1/3 the force (##F_{thrust} = -\Sum q E##)
    What's the draw to the equilateral triangle?
    Perhaps it's just an artifact of my inconsistent definition of r? But it's ratios, and I just used r as a means to calculate and compare, hmm...
  2. jcsd
  3. Jun 15, 2016 #2


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    Staff: Mentor

    Triangles are structurally simple and strong.
  4. Jun 16, 2016 #3

    jack action

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    Gold Member

    It seems the weight is more based on the number of corner since most of the weight seems to be concentrated on the vertical posts.

    The triangle is therefore the practical shape that have the least number of corners.
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