Involute Math

Discussion in 'Research and Development' started by GregMcFadden, Feb 20, 2009.

  1. GregMcFadden

    GregMcFadden Facilitator RCWC Staff

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    Alright folks, here is the challenge of the evening... assuming that it even has a discrete solution... pardon any mistypeing... I keep coming up with one too few equations to solve (one too many unknowns).... which either means I have missed something or there is no unique solution...
    All angles are in radians...

    The goal (pump related, of course)... imagine two circles, an inner circle and an outer circle, centered on location (0,0) in the Primary coordinate system.
    The parametric equations for the circle are:
    x=r*cos(theta)
    y=r*sin(theta)
    r=radius of the circle
    theta is a angle measured from the X axis, in the standard Right hand rule manner
    x & y are the coordinates of the circle for a given theta....
    the tangent angle of the circle will be alpha=theta+Pi/2

    Now imagine an involute, who's characteristic circle is centered a distance XX,YY from (0,0).
    The parametric equation for the involute is a function of the radius of its characteristic circle, and the tangent angle of the curve.... with the origin at the center of the characteristic circle....
    x=a*(cos(t)+t*sin(t))= F(a,t)
    y=a*(sin(t)-t*cos(t[I]))= G(a,t[I])
    where a= the radius of the characteristic circle of the involute curve
    t[I] is the tangent angle of said curve....
    First, one must translate the involute equation into the same coordinate system as the two circles...
    X[I]=x[I] + XX
    Y[I]=y[I] + YY

    [B]Now here is the fun stuff.... [/B]
    Imagine you know the intersection angle of the circles and the involute... E.G. the outer circle and the involute intersect producing an intersection angle of A1. the inner circle and the involute intersect producing an intersection angle of A2.
    in that case.
    A1=theta+Pi/2 - t at the location of the intersection at the outer circle
    A2 = theta + Pi/2 -t at the location of the intersection of the inner circle.
    If you match the parametric equations at both intersecting points, this leaves you 5 equations that fully define point 1, and 5 more that define the other intersection point... all 10 equations are independent...
    There is a problem however... there are 11 unknowns... which means I need another equation (or there is no single solution) So I figured I would give you folks a chance to mull it over...


    [/I][/I][/I][/I][/I][/I][/I]
     
  2. JustinScott

    JustinScott Well-Known Member

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    Not to sound like an invalid.. but what is an invalute?

    Can we get some photos here? I'm pretty good at this stuff (math degree...) but I need to visualize what we are doing & equations alone suck.
     
  3. GregMcFadden

    GregMcFadden Facilitator RCWC Staff

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  4. GregMcFadden

    GregMcFadden Facilitator RCWC Staff

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    something like this although I can't draw an involute to save my life.. the two spots marked with angles are what I want to define the involute by, assuming that is doable... I do not know their X,Y positions.... but I do know the intersection angle I want... I can not simply fix one of them at a point on the circle without allowing the involute's coordinate system to rotate with respect to the circles coordinate system. Luckily, if I don't fix that point, I don't have to deal with rotating coordinate system transformations.. and all this is a skosh fuzzy already.

    Like I said before, there may not be a unique solution, there may even be a set of unique solutions, but operating under the "I need as many equations as I got variables" assumption.... tells me I need one more thing to make this solvable...


    [​IMG][​IMG]
     
  5. squires

    squires Member

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    may i ask the point?? it is possible to solve an equation without as many equations as variables although i think it takes in the billions or iterations for 3 unknowns so for 10 i would imagine it is not really possible unless you have super computer available. the other option is there any change that a variable is named twice. i havent really looked at the problem too much effort.
     
  6. GregMcFadden

    GregMcFadden Facilitator RCWC Staff

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    because I am interested and am trying to figure it out...

    as far as solving, that is the least of my concerns.. finding the last equation I need... would let the computer solve it in seconds, without the other bound or at least an idea where it might be, and there is no single solution... You are greatly underestimating the performance of modern computing systems... I believe you may be confusing unknowns/equatiuons with dimensions... for example, a 3 dimensional finite element analysis problem that will solve in minutes on a normal computer can have hundreds of thousands of equations and unknowns....
     
  7. squires

    squires Member

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    fair enough i thought there was some practical application that you had thought of for it, im sure there indead will be in the future.
     
  8. GregMcFadden

    GregMcFadden Facilitator RCWC Staff

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    It is mostly for pumps, as many components tend towards spirals/involutes in pumps.
     
  9. phill

    phill Active Member

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    I'm not up to the math but I am assuming that what you are trying to do is to create a shape that will tend to a constant increase in the acceleration of the water from when it first encounters the vane until it reaches the tip of the vane. If this is so, then would it make sense to look at the distance from the inter side of the vane to the out side of the circle and how it changes as you move over the arc? I don't quite see whay the angle on the outside needs to be 30 degrees... Isn't the important thing what is happening on the other side?
    phill the semi-clueless
     
  10. GregMcFadden

    GregMcFadden Facilitator RCWC Staff

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    There is a deal of industry data and correlations for setting inlet/outlet angles for various purposes....
     
  11. squires

    squires Member

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    if your really stuck e-mail it to a professor of maths in your local uni they may if your lucky spend a few hours of well paid time(student tuition) looking at it and give you a solution