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Calculate forces and size parts in a steering arrangement

  1. Feb 4, 2010 #1
    1. The problem statement, all variables and given/known data

    I am trying to size parts of a boat, but I don't believe in my results. An attempt to solve one subproblem of this in a rulebook based way can be read here: http://www.boatdesign.net/forums/boat-design/rudder-scantling-31263.html Is it appropriate to bring that question here without modifications, as no one have answered it there?

    There is a steering arrangement of a boat in the drawing below. Force and torsion on the rudder axlle (axle 1) [tex] F_{1} [/tex] and [tex]M_{1}[/tex] are known.

    Calculate the forces on the wire ([tex]Fw_{2}[/tex]), the ring ([tex]F_{4}[/tex]), and the tiller axle ([tex]F_{2}[/tex]).

    Axle 1 and axle 2 are made of circular steel rods, arm made of rectangular cross section plywood 4mm thick, placed horizontally.

    Calculate the radius of axles, and the width of the arm.

    This attempt actually stops at radius of axle 1.

    attachment.php?attachmentid=23521&stc=1&d=1265272201.png
    [tex]$
    known values: \\
    F_{1} = 476 N \\
    M_{1} = 15 N m \\
    l_{1} = 0.1 m \\
    l_{2} = 0.56 m \\
    l_{3} = 0.8 m \\
    l_{4} = 0.05 m \\
    l_{5} = 0.2 m \\
    l_{6} = 0.4 m \\
    [/tex]
    2. Relevant equations

    [tex]\sum F = 0 [/tex]
    [tex]\sum M = 0 [/tex]
    [tex]M = F * l [/tex]

    Euler-Bernoulli beam equation:
    [tex]\sigma = \frac{M y}{Ix}[/tex]
    Second moment of inertia in a circular beam:
    [tex]Ix = \frac{1}{4} \pi r^{4} [/tex]


    3. The attempt at a solution
    [tex]
    $ unit vector in direction for$ Fw_{2} \\
    e_{Fw2}=\left(\begin{smallmatrix}\frac{l_{2}}{\sqrt{l_{1}^{2} + l_{2}^{2}}} & - \frac{l_{1}}{\sqrt{l_{1}^{2} + l_{2}^{2}}}\end{smallmatrix}\right)=\left(\begin{smallmatrix}0.984427575508482 & -0.175790638483658\end{smallmatrix}\right) \\
    \lvert{e_{Fw2}}\rvert=1.0 \\
    $ force in wire 2$ \\
    Fw_{2}=- \frac{M_{1} e_{Fw2}}{l_{1}}=\left(\begin{smallmatrix}147.664136326272 N & - 26.3685957725486 N\end{smallmatrix}\right) \\
    \lvert{Fw_{2}}\rvert=150.0 N \\
    [/tex]
    [tex] $ \\
    unit vector in direction for wire 2 at tiller end $ \\
    e_{Fw2 t}=\left(\begin{smallmatrix}- \frac{l_{1}}{\sqrt{l_{1}^{2} + l_{3}^{2}}} & \frac{l_{3}}{\sqrt{l_{1}^{2} + l_{3}^{2}}}\end{smallmatrix}\right)=\left(\begin{smallmatrix}-0.124034734589208 & 0.992277876713668\end{smallmatrix}\right) \\
    \lvert{e_{Fw2 t}}\rvert=1.0 \\
    $ force in wire 2 at tiller end $ \\
    Fw_{2t}=e_{Fw2 t} \lvert{Fw_{2}}\rvert=\left(\begin{smallmatrix}- 18.6052101883813 N & 148.84168150705 N\end{smallmatrix}\right) \\
    \lvert{Fw_{2t}}\rvert=150.0 N \\
    $ force at ring $ \\
    F_{4}=- Fw_{2} - Fw_{2t}=\left(\begin{smallmatrix}- 129.058926137891 N & - 122.473085734502 N\end{smallmatrix}\right) \\
    \lvert{F_{4}}\rvert=177.920946336276 N \\
    [/tex]
    [tex] $ \\
    $ equation for $ F_{3} $ using moment at $ F_{2} \\
    F_{3} l_{6} - Fw_{2t} l_{1} = 0 \\
    $ steering force at tiller $ \\
    F_{3}=\frac{Fw_{2t} l_{1}}{l_{6}}=\left(\begin{smallmatrix}- 4.65130254709532 N & 37.2104203767625 N\end{smallmatrix}\right) \\
    \lvert{F_{3}}\rvert=37.5 N \\
    $ equation for $ F_{2} $ using moment at joint of tiller and wire 2 \\
    F_{3} \left(l_{1} + l_{6}\right) - F_{2} l_{1} = 0 \\
    $ force on axle 2 $ \\
    F_{2}=\frac{F_{3} l_{1} + F_{3} l_{6}}{l_{1}}=\left(\begin{smallmatrix}- 23.2565127354766 N & 186.052101883813 N\end{smallmatrix}\right) \\
    \lvert{F_{2}}\rvert=187.5 N \\
    F_{2}=\frac{F_{3} l_{1} + F_{3} l_{6}}{l_{1}}=\left(\begin{smallmatrix}- 23.2565127354766 N & 186.052101883813 N\end{smallmatrix}\right) \\
    \lvert{F_{2}}\rvert=187.5 N \\
    $ Fb_{1} = Fb_{2} = F_{1}/2 \\
    Fb_{1}=\frac{1}{2} F_{1}=238 N \\
    \lvert{Fb_{1}}\rvert=238 N \\
    Fb_{2}=\frac{1}{2} F_{1}=238 N \\
    \lvert{Fb_{2}}\rvert=238 N \\
    [/tex]
    [tex] $ \\

    shear in axle 1 $ \\
    shear = \begin{cases} 0 & \text{for}\: x < 0 \\Fb_{1} & \text{for}\: \operatorname{And}\left(0 \leq x,x < l_{5}\right) \\Fb_{1} - F_{1} & \text{for}\: \operatorname{And}\left(l_{5} \leq x,x < 2 l_{5},0 \leq x\right) \\\operatorname{And}\left(2 l_{5} \leq x,l_{5} \leq x,0 \leq x\right) & \text{for}\: Fb_{1} + Fb_{2} - F_{1} \end{cases} \\
    moment = \begin{cases} 0 & \text{for}\: x < 0 \\Fb_{1} x & \text{for}\: \operatorname{And}\left(0 \leq x,x < 0.2 m\right) \\x \left(Fb_{1} - F_{1}\right) + 0.2 F_{1} m & \text{for}\: \operatorname{And}\left(0.2 m \leq x,x < 0.4 m\right) \\0.4 m \leq x & \text{for}\: x \left(Fb_{1} + Fb_{2} - F_{1}\right) + 0.2 F_{1} m - 0.4 Fb_{2} m \end{cases} \\
    [/tex]
    attachment.php?attachmentid=23520&stc=1&d=1265266152.png

    [tex] $ \\
    maximum bending moment in axle 1 $ \\
    M_{a1}=Fb_{1} l_{5}=47.6 N m \\
    \lvert{M_{a1}}\rvert=47.6 N m \\
    [/tex]
    [tex] \\
    $ Euler-Bernoulli beam equation: $ \
    \sigma = \frac{M y}{Ix} \\
    $ second moment of inertia in a circular beam: $ \
    Ix = \frac{1}{4} \pi r^{4} \\
    $ substituing the two equations together, $ r = y $ and $ M = M_{a1} \\
    \sigma = 4 \frac{M_{a1}}{\pi r^{3}} \\
    $ solved for r :$ \\
    r = \frac{2^{\frac{2}{3}} \sqrt[3]{M_{a1}} \sqrt[3]{\frac{1}{\sigma}}}{\sqrt[3]{\pi}} \\

    [/tex]

    Now I have a major problem here. I don't know which value to substitute for [tex]\sigma[/tex].

    Looking up physical properties of materials, the ones expressed in [tex]\frac{N}{m^{2}}[/tex] are:

    Ultimate tensile strength
    Yielding tensile strength
    Modulus of Elasticity (I think I understand that one)
    Compressive yield strength
    Bulk modulus
    Fatique strength
    Shear modulus
    Shear strength
    Flexural modulus

    I bet at yielding tensile strength, but not sure.

    I calculate with 200Mpa

    [tex]
    r=\frac{2^{\frac{2}{3}} \sqrt[3]{M_{a1}} \sqrt[3]{\frac{1}{\sigma}}}{\sqrt[3]{\pi}}=0.247448994460805 m 2^{\frac{2}{3}} \sqrt[3]{5} \\
    \lvert{r}\rvert=0.247448994460805 m 2^{\frac{2}{3}} \sqrt[3]{5} \\
    r= 0.671679909773 \\
    [/tex]
    I found this value rather off the scale, so there should be something wrong with my calculations.
     

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    Last edited: Feb 4, 2010
  2. jcsd
  3. Feb 6, 2010 #2
    Bumping up, full with hope.
     
  4. Feb 6, 2010 #3

    nvn

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    magwas: This is a school assignment, right? You have a correct relevant equation, M = L*F, where L = perpendicular distance to the force. But if you want to use vectors, you need to use M = r X F, not dot product (or whatever you used, which looks incorrect). Therefore, your first mistake is your formula and answer for Fw2. Try again. Also, we do not know where L6 is located, because it is not shown in your diagram. Also, is the green wire in your diagram in the horizontal plane?
     
  5. Feb 7, 2010 #4
    Thank you for your help.

    Not actually school assignment, as I am trying to learn it for myself, but it does not matter, I think.

    I have uploaded an updated picture.
    This time I tried to use scalar forces with moment calculations. I just can't get the vector part right.
    Tried to play with [tex] F3 = \frac{\lvert{Fw_{2t}}\rvert \lvert{l_{1}}\rvert \lvert{l_{3}}\rvert}{\sqrt{l_{1}^{2} + l_{3}^{2}}}[/tex], but I am no closer.

    Also because my math package have some problems with vectors, I am using complex numbers for the remaining vector calculations.
    [tex]

    $ b is length of (p1 axle1), a is length of (p1 ring) $ \\
    $ such that (p1 axle1) is normal to Fw2 $ \\
    $ now $ \frac{l_{1}}{l_{2}} = \frac{a}{c} and l_{2}^{2} = a^{2} + c^{2} \\
    c = \frac{a l_{2}}{l_{1}} $ so $ l_{2}^{2} = a^{2} + \frac{a^{2} l_{2}^{2}}{l_{1}^{2}} \\
    a=\frac{\lvert{l_{1}}\rvert \lvert{l_{2}}\rvert}{\sqrt{l_{1}^{2} + l_{2}^{2}}}=0.0984427575508482 m \\
    \lvert{a}\rvert=0.0984427575508482 m \\
    [/tex]
    [tex]
    $ unit vector in direction for$ Fw_{2} \\
    e_{Fw2}=\frac{l_{2} - \mathbf{\imath} l_{1}}{\sqrt{l_{1}^{2} + l_{2}^{2}}}=1.75790638483657 \frac{0.56 m - 0.1 \mathbf{\imath} m}{m} \\
    \lvert{e_{Fw2}}\rvert=1.0 \\
    $ Moment $ \
    M_{1} = Fw_{2} a \\
    $ force in wire 2$ \\
    Fw_{2}=\frac{M_{1} e_{Fw2}}{a}=267.857142857143 \frac{N \left(0.56 m - 0.1 \mathbf{\imath} m\right)}{m} \\
    \lvert{Fw_{2}}\rvert=152.372814142799 N \\
    [/tex]
    [tex] $ \\
    unit vector in direction for wire 2 at tiller end $ \\
    e_{Fw2 t}=\frac{- l_{1} + \mathbf{\imath} l_{3}}{\sqrt{l_{1}^{2} + l_{3}^{2}}}=1.24034734589208 \frac{- 0.1 m + 0.8 \mathbf{\imath} m}{m} \\
    \lvert{e_{Fw2 t}}\rvert=1.0 \\
    $ force in wire 2 at tiller end$ \\
    Fw_{2t}=e_{Fw2 t} \lvert{Fw_{2}}\rvert=188.995215608128 \frac{N \left(- 0.1 m + 0.8 \mathbf{\imath} m\right)}{m} \\
    \lvert{Fw_{2t}}\rvert=152.372814142799 N \\
    $ force at ring $ \\
    F_{4}=- Fw_{2} - Fw_{2t}=- 188.995215608128 \frac{N \left(- 0.1 m + 0.8 \mathbf{\imath} m\right)}{m} - 267.857142857143 \frac{N \left(0.56 m - 0.1 \mathbf{\imath} m\right)}{m} \\
    \lvert{F_{4}}\rvert=180.735435254722 N \\
    [/tex]
    [tex] $ \\
    $ equation for $ F_{3} $ using moment at $ F_{2} \\
    F_{3} l_{6} - \frac{\lvert{Fw_{2t}}\rvert \lvert{l_{1}}\rvert \lvert{l_{3}}\rvert}{\sqrt{l_{1}^{2} + l_{3}^{2}}} = 0 \\
    $ steering force at tiller $ \\
    F_{3}=\frac{\lvert{Fw_{2t}}\rvert \lvert{l_{1}}\rvert \lvert{l_{3}}\rvert}{l_{6} \sqrt{l_{1}^{2} + l_{3}^{2}}}=37.7990431216257 N \\
    \lvert{F_{3}}\rvert=37.7990431216257 N \\
    $ equation for $ F_{2} $ using moment at joint of tiller and wire 2 \\
    F_{3} \left(l_{1} + l_{6}\right) - F_{2} l_{1} = 0 \\
    $ force on axle 2 $ \\
    F_{2}=\frac{F_{3} l_{1} + F_{3} l_{6}}{l_{1}}=188.995215608128 N \\
    \lvert{F_{2}}\rvert=188.995215608128 N \\
    F_{2}=\frac{F_{3} l_{1} + F_{3} l_{6}}{l_{1}}=188.995215608128 N \\
    \lvert{F_{2}}\rvert=188.995215608128 N \\
    [/tex]


    Is the above correct now?

    I think I will ask the second part in a separate question, because it is ... a separate question.
     

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    Last edited: Feb 7, 2010
  6. Feb 7, 2010 #5

    nvn

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    magwas: Your answer for Fw2, Fw2t, F4, and F3 is now correct. F2 is currently incorrect due to a mistake in your moment summation equation for F2. In that equation, change (L1 + L6) to (L6 - L1). Also, is the green wire in your diagram in the horizontal plane, parallel to the water surface?
     
  7. Feb 7, 2010 #6
    Thank you for your help again.

    Yes, the green wires are horizontal.

    The relevant part now:
    [tex]
    $ equation for $ F_{2} $ using moment at joint of tiller and wire 2 \\
    F_{3} \left(l_{6} - l_{1}\right) - F_{2} l_{1} = 0 \\
    $ force on axle 2 $ \\
    F_{2}=\frac{F_{3} l_{6} - F_{3} l_{1}}{l_{1}}=113.397129364877 N \\
    \lvert{F_{2}}\rvert=113.397129364877 N \\
    F_{2}=\frac{F_{3} l_{6} - F_{3} l_{1}}{l_{1}}=113.397129364877 N \\
    \lvert{F_{2}}\rvert=113.397129364877 N \\
    [/tex]

    I think I can go further from this point.
    The plan: As the arm rotates, the angles will change. I think I will set a limit at 60 degrees, which approximately doubles the forces in the wire, Also it is just half of the steering arrangement (it is symmetric to drawn angle of tiller), which means a force in the mirror of wire 1, approximately the same magnitude as Fw2, so F2 will be doubled again. I will calculate with 320N for Fw2, and 640N for F2 to be on the safe side.
     
  8. Feb 7, 2010 #7

    nvn

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    magwas: F2 is now correct. In your last paragraph, do you mean there is a mirror of the system shown in your diagram, and the mirror system is omitted from your diagram? If so, then if the 60 deg angle (where is the angle?) approximately doubles the wire tensile force, then this would quadruple F3, giving F3 = 151.2 N. But notice this would change F2 to 151.2 N, not 640 N.

    Whenever you draw a free-body diagram, draw the entire system or subsystem, showing all forces acting on the system or subsystem. You cannot draw only some of the forces acting on the subsystem, and omit other significant forces acting on the subsystem. Do you have only one tiller? Draw a correct free-body diagram of the tiller, showing all forces acting on it, including forces from the port (mirror) steering system.
     
  9. Feb 7, 2010 #8

    nvn

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    In the meantime, until you come back to post 7, let's move on to the remainder of what you posted in post 1. Your answer for Fb1 and Fb2 is incorrect. If F1 = 476 N, then summation of moment, and summation of horizontal forces, on axle 1 gives Fb1 = 595 N, and Fb2 = F1 - Fb1 = -119 N.

    Your shear diagram does not make sense, because it uses the wrong x coordinates. From the data given earlier in post 1, we know L4 = 0.05 m, and L5 = 0.2 m. Therefore, if x = 0 at F1, then your shear diagram should be V(x) = {F1 for x ≤ L4; and F1 - Fb1 for x > L4}. And your moment diagram should be M(x) = {F1*x for x ≤ L4; and F1*x - Fb1*(x - L4) for x > L4}. Therefore, the maximum bending moment in axle 1 is Ma1 = 23.8 N*m = 23 800 N*mm.

    Let's say, e.g., your axle 1 material tensile yield strength is Sty = 235 MPa. You could perhaps use a yield factor of safety of FSy = 1.70. Therefore, the required axle 1 radius becomes r1 = [4*Ma1*FSy/(pi*Sty)]^0.33333 = {4(23 800 N*mm)*1.70/[pi*(235 MPa)]}^0.33333 = 6.030 mm.

    Now repeat the above example for when the steering system is rotated to your 60 deg rotation limit (if rotating the steering system increases F1).
     
  10. Feb 8, 2010 #9
    Thank you for your neverending patience.
    I have inserted a "2*" into the calculation of Fw2, and another "2*" into the equation for F3.
    The 60 degrees meant to be rotation of arm around axle 1 by 60 degrees. If the wires would be paralell to axle 1 - ring, then the perpendicular distance would be l1/2 in this case. Because they are not paralell, the distance is a bit less when the direction of rotation is clockwise. I neglect this because 60 degrees is just a safe bet, and M1 also contains appropriate safety margins.
    The "2*" in the equation for F3 needs some explanation, exactly because I did not draw the whole steering arangement up.
    Correction:
    attachment.php?attachmentid=23592&stc=1&d=1265616238.jpg

    The mirror part is mirrored, but direction of M1 is the same, so forces are pulling mirror of wire 1, not wire 2.
    This goes to the tiller at the end, same magnitude and direction as caused by Fw2. Thence the "2*"

    Now from that point on:
    [tex]
    $ force in wire 2$ \\
    Fw_{2}=2 \frac{M_{1} e_{Fw2}}{a}=535.714285714286 \frac{N \left(0.56 m - 0.1 \mathbf{\imath} m\right)}{m} \\
    \lvert{Fw_{2}}\rvert=304.745628285598 N \\
    [/tex]
    [tex] $ \\
    unit vector in direction for wire 2 at tiller end $ \\
    e_{Fw2 t}=\frac{- l_{1} + \mathbf{\imath} l_{3}}{\sqrt{l_{1}^{2} + l_{3}^{2}}}=1.24034734589208 \frac{- 0.1 m + 0.8 \mathbf{\imath} m}{m} \\
    \lvert{e_{Fw2 t}}\rvert=1.0 \\
    $ force in wire 2 at tiller end$ \\
    Fw_{2t}=e_{Fw2 t} \lvert{Fw_{2}}\rvert=377.990431216257 \frac{N \left(- 0.1 m + 0.8 \mathbf{\imath} m\right)}{m} \\
    \lvert{Fw_{2t}}\rvert=304.745628285598 N \\
    $ force at ring $ \\
    F_{4}=- Fw_{2} - Fw_{2t}=- 377.990431216257 \frac{N \left(- 0.1 m + 0.8 \mathbf{\imath} m\right)}{m} - 535.714285714286 \frac{N \left(0.56 m - 0.1 \mathbf{\imath} m\right)}{m} \\
    \lvert{F_{4}}\rvert=361.470870509445 N \\
    [/tex]

    equation for F3 using moments at F2, featuring moment by wire 1 as "2*"
    [tex]
    F_{3} l_{6} - 2 \frac{\lvert{Fw_{2t}}\rvert \lvert{l_{1}}\rvert \lvert{l_{3}}\rvert}{\sqrt{l_{1}^{2} + l_{3}^{2}}} = 0
    [/tex]
    [tex] $ \\
    $ steering force at tiller $ \\
    F_{3}=2 \frac{\lvert{Fw_{2t}}\rvert \lvert{l_{1}}\rvert \lvert{l_{3}}\rvert}{l_{6} \sqrt{l_{1}^{2} + l_{3}^{2}}}=151.196172486503 N \\
    \lvert{F_{3}}\rvert=151.196172486503 N \\
    [/tex]
    [tex] $ \\
    $ equation for $ F_{2} $ using moment at joint of tiller and wire 2 \\
    F_{3} \left(l_{6} - l_{1}\right) - F_{2} l_{1} = 0 \\
    $ force on axle 2 $ \\
    F_{2}=\frac{F_{3} l_{6} - F_{3} l_{1}}{l_{1}}=453.588517459508 N \\
    \lvert{F_{2}}\rvert=453.588517459508 N \\
    F_{2}=\frac{F_{3} l_{6} - F_{3} l_{1}}{l_{1}}=453.588517459508 N \\
    \lvert{F_{2}}\rvert=453.588517459508 N \\
    [/tex]

    I should really make a separate question about sizing axle 1, because it is more complex. There is not just F1, but also F2 and M1 are acting on the axle.
    On the first drawing, I have given the attack point of F2 on axle 1 incorrectly, but my attempt at calculation tried to use the correct one, maybe incorrectly. I see that I could correctly bet on the type of tension, the formula for r was nearly okay, but I forgot to apply a safety factor and maybe there were problems with my shear and moment calculation. I will ask that part in a separate thread sometimes afternoon CET, and make a note of it here.

    And thank you again.
     

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    Last edited: Feb 8, 2010
  11. Feb 8, 2010 #10
  12. Feb 8, 2010 #11

    nvn

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    magwas: What is the forward direction of the boat? Show the forward direction of motion on your diagram. Is it to the left? Perhaps define x, y, and z axes on your diagram(s). Let +z be upward, perpendicular to the water surface. Let's say angle theta is the rudder rotation angle. We do not know if theta is measured from a horizontal or vertical line on your diagram. Show theta on your diagram. And show which direction is positive theta. Positive theta is often counterclockwise. Why are the rudders perpendicular to the boat forward direction of motion? Shouldn't theta = 0 deg be when the rudders are parallel to the forward direction of motion of the boat? And therefore, your current diagram shows the rudders at theta = 90 deg, correct? Or does your diagram currently show theta = -90 deg?

    Can you show a good diagram of your rudder, with the rudder parts? We need to see the rudder assembly to be able to see if you are applying the applied loads correctly, and to ensure the rudder forces and moments are in equilibrium.

    The rudder forces and equations need to be set up for when theta is at an arbitrary value (not 0 nor 90 deg), so that the analysis will give the correct forces, in equilibrium, at any angle theta.
     
  13. Feb 8, 2010 #12

    nvn

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    magwas: If the rudder in your above diagram is currently shown at theta = 0 deg, and if M1 = 15 N*m is truly correct and constant as the rudder rotates, and if you rotate the rudder counterclockwise to theta = +60 deg, then the angle of starboard wire 2, measured from a horizontal line in your diagram, decreases to phi = 6.0292 deg. The wire angle does not remain constant as the rudder rotates; it changes. If you work out the trigonometry and vector mechanics correctly, you will see Fw2 = 255.02 N, not 304.75 N, if the rudder is rotated 60 deg counterclockwise from the rudder position currently shown in your above diagram.

    Also, your moment summation equation for F2 is currently incorrect. You forgot to apply the port wire 1 tensile force to the end of the tiller. Therefore, your moment summation equation for F2 is wrong. Hence, your answer for F2 is incorrect.

    First, define theta, as mentioned in post 11. Also, define phi.
     
  14. Feb 9, 2010 #13
    You can see the whole boat here: http://www.cruiserlog.com/forums/index.php?act=attach&type=post&id=3007
    The rudder itself is not shown in the diagram, it is right to the arms, 90 degrees. Theta is 0 degrees in the diagram, this is the case when the boat goes forward.
    I believe I have applied port wire 1 to the tiller. See the "2*" in the expression after the comment "featuring moment by wire 1 as "2*""

    This state of affairs would be okay for me, as I believe I have established safe upper limits for the forces, but to improve my engineering abilities I will do the drawing and computations as you have suggested. I hope I can come back with it later today.
     
  15. Feb 9, 2010 #14

    nvn

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    Yes, the "2*" causes F3 to be correct. I do not have an issue with your F3 answer. As mentioned in paragraph 2 of post 12, the mistake is with your last moment summation equation in post 9, to solve for F2. You have not created a correct free-body diagram of the tiller to solve for F2. And therefore, your answer for F2 is incorrect.
    Agreed. Your assumption (overestimation) of Fw2 = 304.75 N causes the axle 1 radius to be only 1.3 % too big, which is OK.
     
  16. Feb 10, 2010 #15
    Thank you again.
    Now, this is just about what I can do with this.
    I was struggling with perpendicular distance, cross product and dot product for a day, now I believe I have something useable for F2.
    A new - hopefully more understandable - drawing:
    attachment.php?attachmentid=23636&stc=1&d=1265811753.jpg

    [tex]armend=e_{arm} l_{1}[/tex]
    unit vector in direction for [tex]Fw_{2}[/tex]
    [tex]e_{Fw2}=\frac{armend + l_{2}}{\lvert{armend + l_{2}}\rvert}[/tex]
    Moment [tex]M_{1} = Fw_{2} \left(\Im{armend} \Re{e_{Fw2}} - \Im{e_{Fw2}} \Re{armend}\right)[/tex]
    force in wire 2
    [tex]Fw_{2}=\frac{M_{1}}{\Im{armend} \Re{e_{Fw2}} - \Im{e_{Fw2}} \Re{armend}}[/tex]
    [tex]Fw_{1}=Fw_{2}[/tex]
    unit vector in direction for wire 2 at tiller end
    wire2 vector: [tex] - \mathbf{\imath} l_{3} + \mathbf{\imath} e_{arm} l_{1} [/tex]. its unit vector:
    [tex]e_{Fw2 t}=\frac{- \mathbf{\imath} l_{3} + \mathbf{\imath} e_{arm} l_{1}}{\lvert{- \mathbf{\imath} l_{3} + \mathbf{\imath} e_{arm} l_{1}}\rvert}[/tex]
    unit vector in direction for wire 1 at tiller end
    wire2 vector: [tex] - \mathbf{\imath} l_{3} - \mathbf{\imath} e_{arm} l_{1} [/tex]. its unit vector:
    [tex]e_{Fw1 t}=\frac{- \mathbf{\imath} l_{3} - \mathbf{\imath} e_{arm} l_{1}}{\lvert{- \mathbf{\imath} l_{3} - \mathbf{\imath} e_{arm} l_{1}}\rvert}[/tex]
    force in wire 2 at tiller end
    [tex]Fw_{t 2}=e_{Fw2 t} \lvert{Fw_{2}}\rvert[/tex]
    force in wire 1 at tiller end
    [tex]Fw_{t 1}=e_{Fw1 t} \lvert{Fw_{1}}\rvert[/tex]
    force at ring, starboard side
    [tex]F_{4}=- Fw_{t 2} - Fw_{2} e_{Fw2}[/tex]
    force at ring, port side
    [tex]F_{p 4}=- Fw_{t 1} - Fw_{1} e_{Fw1}[/tex]
    equations for F2 using moments at F_3
    eq1=F2_re*l6+Fw2*(l6-abs(perpdist(l3*1j,0,e_Fw2_t)))-Fw1*(l6+abs(perpdist(-1j*l3,0,e_Fw1_t)))
    eq2=F2_im+dot(Fw2*e_Fw2_t,1j*e_arm)-dot(Fw1*e_Fw1_t,1j*e_arm)
    eq3=Equality(F2,F2_re*1j*e_arm+F2_im*e_arm)

    steering force at tiller
    [tex]F_{2}=\frac{\mathbf{\imath} Fw_{1} e_{arm} l_{6} + \mathbf{\imath} Fw_{1} e_{arm} \lvert{\frac{l_{3} \Re{e_{Fw1 t}}}{\left(\Im{e_{Fw1 t}}\right)^{2} + \left(\Re{e_{Fw1 t}}\right)^{2}}}\rvert + \mathbf{\imath} Fw_{2} e_{arm} \lvert{\frac{l_{3} \Re{e_{Fw2 t}}}{\left(\Im{e_{Fw2 t}}\right)^{2} + \left(\Re{e_{Fw2 t}}\right)^{2}}}\rvert + e_{arm} l_{6} \Im{e_{arm}} \Re\left(Fw_{2} e_{Fw2 t}\right) + e_{arm} l_{6} \Im\left(Fw_{1} e_{Fw1 t}\right) \Re{e_{arm}} - \mathbf{\imath} Fw_{2} e_{arm} l_{6} - e_{arm} l_{6} \Im{e_{arm}} \Re\left(Fw_{1} e_{Fw1 t}\right) - e_{arm} l_{6} \Im\left(Fw_{2} e_{Fw2 t}\right) \Re{e_{arm}}}{l_{6}}[/tex]
    at alpha=0 degrees
    force in wire 1
    [tex]Fw_{1}=152.372814142799 N[/tex]
    [tex]\lvert{Fw_{1}}\rvert=152.372814142799 N[/tex]
    [tex]Fw_{2}=152.372814142799 N[/tex]
    [tex]\lvert{Fw_{2}}\rvert=152.372814142799 N[/tex]
    [tex]F_{4}=- 131.100478439187 N + 124.410458200788 \mathbf{\imath} N[/tex]
    [tex]\lvert{F_{4}}\rvert=180.735435254722 N[/tex]
    [tex]F_{p 4}=- 168.899521560813 N + 124.410458200788 \mathbf{\imath} N[/tex]
    [tex]\lvert{F_{p 4}}\rvert=209.773712588593 N[/tex]
    [tex]F_{2}=- 75.5980862432514 N - 37.7990431216257 \mathbf{\imath} N[/tex]
    [tex]\lvert{F_{2}}\rvert=84.5212299044009 N[/tex]
    at alpha=60 degrees
    force in wire 1
    [tex]Fw_{1}=- 151.311909944752 N[/tex]
    [tex]\lvert{Fw_{1}}\rvert=151.311909944752 N[/tex]
    [tex]Fw_{2}=- 151.311909944752 N[/tex]
    [tex]\lvert{Fw_{2}}\rvert=151.311909944752 N[/tex]
    [tex]F_{4}=131.682426318086 N + 123.540878611267 \mathbf{\imath} N[/tex]
    [tex]\lvert{F_{4}}\rvert=180.561928681165 N[/tex]
    [tex]F_{p 4}=167.507853154076 N + 123.380418154109 \mathbf{\imath} N[/tex]
    [tex]\lvert{F_{p 4}}\rvert=208.042323704025 N[/tex]
    [tex]F_{2}=- 78.6564331483069 N - 10.703545171187 \mathbf{\imath} N[/tex]
    [tex]\lvert{F_{2}}\rvert=2.5 \sqrt{1008.22405677531 N^{2} - 2.8421709430404 \times 10^{-14} \mathbf{\imath} N^{2}}[/tex]
    Fw1 blue, F4 green, F4 port red, F2 cyan
    attachment.php?attachmentid=23637&stc=1&d=1265811753.png

    i hereby declare this example to be beaten to death.
     

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  17. Feb 10, 2010 #16

    nvn

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    For M1 = 15 N*m and alpha = 60 deg, Fw1 = 255.02 N, not 151.32 N. When you work the problem correctly, F2 = -F3. For alpha = 60 deg, F2 = -69.87 N.
     
  18. Feb 11, 2010 #17
    Thank you again.

    Reworked the equations, and double-checked them for problems with sign, etc.
    Here is the result:
    attachment.php?attachmentid=23670&stc=1&d=1265930988.jpg

    I am back to basics, analised that Fw2 force and the related angles.
    In the image
    - green is beta, the angle of e_Fw2 as calculated
    - blue is 90+alpha, that is the ring-axle1-p1 angle
    - red is 90-alpha-beta, that is ring-p1-axle1 angle
    - the bottom of the double line is 150*1/sin(red)
    - the top of the double line is -l_Fw2
    I am starting to believe my results now, but they are different than the numbers you have given.
    at zero, l_Fw2=-152.37 N
    at 60, it is -347.42 N It is more than double than at zero, because red gets as low as 25 degrees.

    attachment.php?attachmentid=23669&stc=1&d=1265930726.png

    I have realized, that Fw1=Fw2, because the arm is in the same angle.
    so Fw1_t and Fw2_t are mirrors to axle 2.
    Which means that their component paralell to the tiller are cancel each other, and they have the same component orthogonal to the tiller, but in the meaning that their moment around axle2 adds up.
    However their orthogonal component is canceled out by F3. As F3 is by definition orthogonal to the tiller,
    it have no paralell component. So F2 must be zero.
    Or more realistically it is in the direction of the tiller, and its magnitude is the same as the paralell component of F3. So it should be sized according to the strength of the skipper:)

    Where did I err this time?
     

    Attached Files:

  19. Feb 12, 2010 #18

    nvn

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    magwas: Your first mistake is, your wire tensile force for alpha = 60 deg is incorrect. It should instead be Fw1 = 255.02 N.

    L1 = 100 mm, L2 = 560 mm, M1 = 15 000 N*mm.
    At alpha = 60 deg,
    armend = (-86.60254*i - 50*j + 0*k) mm,
    gamma = 6.02920 deg,
    e_Fw1 = -0.994468*i + 0.105035*j + 0*k,
    Fw1 = -M1/(armend X e_Fw1)3 = 255.02 N
     
    Last edited: Feb 12, 2010
  20. Feb 12, 2010 #19
    Thank you again.
    There had been some inconsistencies in my naming scheme. Fw1 refers to the force on the port wire 1.
    You have given e_Fw1 and Fw1 as the direction and magnitude of starboard wire 1, which is the same as port wire 2.
    Starboard wire 1 and port wire 2 have no tension wth this direction of M1.
    (Wire 1 is the inner, wire 2 is the outer wire).
    Here is the drawing in greater resolution.
    attachment.php?attachmentid=23677&stc=1&d=1265963463.png
     

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    Last edited: Feb 12, 2010
  21. Feb 12, 2010 #20

    nvn

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    magwas: In your diagram, alpha is drawn positive clockwise, where alpha = 0 deg means the steering system is in the straight-ahead position. To make a left-hand turn, you move the tiller at F3 toward the starboard side of the boat, correct? I.e. (that is), for positive alpha, you apply a force to the tiller at F3 opposite of the direction currently shown for F3. This causes starboard wire 1 and port wire 2 to be in tension, which rotates the rudder clockwise. Starboard wire 2 and port wire 1 have zero tension. Positive alpha causes the water to apply a counterclockwise moment M1 to the rudder, opposite of the direction currently shown for M1. For positive alpha, M1 is resisted by starboard wire 1 and port wire 2 tension. In other words, you have currently drawn M1 and F3 backwards. And you have currently drawn Fw1 and Fw2 on the wrong wires.

    Another mistake you made in post 17 was saying, "F2 must be zero." That is incorrect. F2 = -F3, as mentioned in post 16.
     
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