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Math Help - Calculate crank angle given pushrod displacement

  1. #1
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    Calculate crank angle given pushrod displacement

    This looks like it should be fairly straightforward on paper, but ...

    Looking to calculate the angle of crank rotation (alpa), about pivot point, given linear displacement of pushrod of 30 units.

    Knowns:

    Pushrod length = 150
    Crank length = 40
    Pushrod displacement = 30

    Thus:


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  2. #2
    MHF Contributor ebaines's Avatar
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    Re: Calculate crank angle given pushrod displacement

    You can proceed as follows, using the dimensions I added to your figure:

    1. The x, y coordinates of the tip of the crank are (R sin alpha, R cos alpha)
    2. Dimension h is R-y
    3. Dimension d is sqrt(R^2-h^2)
    4. Finally, dimension D is L+x-d, or:

     D = L - R \sin \alpha - \sqrt {L^2-R^2 (1-\cos \alpha)^2}

    Now the problem is how to make the angle alpha the independent variable - it's a mess, but after some manipulation I get the folowing quadratic equation:

    (c^2+1)sin^2 \alpha + 2kc \sin \alpha + (k^2+1) = 0

    where c and k are:

     c = \frac {L-D} R

     k = \frac {D^2/2-LD+R^2}{R^2}

    This will yield two values for alpha. Take the arcsin to find the corresponding angles, but as always with arcsins be careful to understand which quadrant alpha belongs in.

    Calculate crank angle given pushrod displacement-crank2.jpg
    Last edited by ebaines; July 22nd 2014 at 10:13 AM.
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    Re: Calculate crank angle given pushrod displacement

    Another way if looking at it: If we set up and xy-coordinate system with origin at the center of the wheel, x axis horizontal and y axis vertical, when the crank arm makes angle \theta with the positive x axis, the end of the crank arm is at (40 cos(\theta), 40 sin(\theta)). The push rod is a straight line of length 150 from [tex](40 cos(\theta), 40 sin(\theta))[tex] to the point (-150+ 30, 0)= (-120, 0). So we have length squared equal to (-120- 40cos(\theta))^2+ (40 sin(\theta))^2= 150^2. That reduces to 9600 cos(\theta)= 2500.
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    MHF Contributor ebaines's Avatar
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    Re: Calculate crank angle given pushrod displacement

    Quote Originally Posted by HallsofIvy View Post
    (-120- 40cos(\theta))^2+ (40 sin(\theta))^2= 150^2.
    I think this should be: (-120- 40cos(\theta))^2+ (40-40 sin(\theta))^2= 150^2

    The answer I get for D=30 is theta = 47.4 degress, or alpha = 90-theta = 42.6 degrees.
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    Re: Calculate crank angle given pushrod displacement

    Thanks for looking into this. I'm still a little lost as neither approach above gets me in the ball park.

    47.4 degrees looks about right for alpha, not theta (drawing this out to scale)?

    Can you please just clarify what R is in your equations above? I assume R=40, but then I'm a little lost with equation (3) Dimension d is sqrt(R^2-h^2)? Is R here not 150?
    Last edited by russiver; July 22nd 2014 at 02:03 PM.
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    MHF Contributor ebaines's Avatar
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    Re: Calculate crank angle given pushrod displacement

    Quote Originally Posted by russiver View Post
    47.4 degrees looks about right for alpha, not theta
    Right - sorry I got that backwards. Shoukld be: alpha = 47.4 degress, or theta= 90-theta = 42.6 degrees.

    Quote Originally Posted by russiver View Post
    Can you please just clarify what R is in your equations above? I assume R=40...
    Yes, R=radius of the circle the crank turns in, or 40. I should have made that clearer. Also L = length of the pushrod =150.

    Quote Originally Posted by russiver View Post
    Dimension d is sqrt(R^2-h^2)? Is R here not 150?
    Sorry, that's another typo. Thanks for catching these errors - I must be getting tired! Equation 3 in my original response should be: Dimension d is sqrt(L^2-h^2).

    By the way, there's another solution: Alpha=169.5 degrees also results in a displacement of 30.
    Last edited by ebaines; July 22nd 2014 at 02:25 PM.
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    Re: Calculate crank angle given pushrod displacement

    With you now. Thanks again for working this through.

    Just need to figure out how you got to (c^2+1)sin^2 \alpha + 2kc \sin \alpha + (k^2+1) = 0 from D = L - R \sin \alpha - \sqrt {L^2-R^2 (1-\cos \alpha)^2}.

    My algebra is pretty rusty, but I'm hoping I can get there. Getting rid of  cos \alpha is my biggest concern. I'm thinking (probably wrong) a trig identity is needed?
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    Re: Calculate crank angle given pushrod displacement

    Quote Originally Posted by ebaines View Post
    I think this should be: (-120- 40cos(\theta))^2+ (40-40 sin(\theta))^2= 150^2
    Yes, I forgot that the center of the wheel was 40 cm below the tip of the pushrod. Thanks.

    The answer I get for D=30 is theta = 47.4 degress, or alpha = 90-theta = 42.6 degrees.
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  9. #9
    MHF Contributor ebaines's Avatar
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    Re: Calculate crank angle given pushrod displacement

    The algebra is messy. First get the square root term by itself, then square both sides. You'll end up with terms of sin^2, sin, and cos and cos^2. You can then replace cos^2 alpha + sin^2 alpaha with 1, and replace the remaining cosine term with sqrt(1-sin^2). Rearrange to get that square root by itself, square both sides, do some final rearrangement, and you're done!
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    Re: Calculate crank angle given pushrod displacement

    Thanks. I''l give this a try and see how I get on.
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    Re: Calculate crank angle given pushrod displacement

    Ok so we start with:

     D = L - R sin \alpha - \sqrt {L^2-R^2 (1- cos \alpha)^2}


    Isolating square root term:

    \sqrt {L^2-R^2 (1- cos \alpha)^2}=  L - Rsin \alpha -D


    Squaring both sides:

    {L^2-R^2 (1- cos \alpha)^2} = L^2 - R^2sin^2\alpha - D^2


    Dividing by R2:

    \frac {L^2} {R^2}-(1- cos \alpha)^2 = \frac {L^2-D^2} {R^2} - sin^2\alpha


    Expanding brackets:

    \frac {L^2} {R^2}- 1 + 2cos\alpha + cos^2 \alpha = \frac {L^2-D^2} {R^2} - sin^2\alpha


    Replacing cos^2 \alpha + sin^2 \alpha with 1:

    \frac {L^2} {R^2}- 1 + 2cos\alpha + 1 = \frac {L^2-D^2} {R^2}


    Am I good so far? Concerned I'm going to lose L?
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    Re: Calculate crank angle given pushrod displacement

    Can already see that I messed up removing the square root - squaring both sides. Apologies algebra and I parted company some years ago!
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  13. #13
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    Re: Calculate crank angle given pushrod displacement

    First - my apologies, but there was another typo on my part. The correct formula is:

    D= L+R \sin \alpha  - \sqrt{L^2-R^2(1-\cos \alpha)^2}

    Rearrange to get:

     L+R \sin \alpha-D= \sqrt{L^2-R^2(1-\cos \alpha)^2}

    and then square both sides. The square of L+R \sin \alpha -D is:

     (L+R \sin \alpha -D)((L+R \sin \alpha -D) = L^2 + R^2 \sin^2 \alpha + D^2 +2LR \sin \alpha -2LD -2DR \siin \alpha

    So we have:

    L^2 + R^2 \sin^2 \alpha + D^2 +2LR \sin \alpha -2LD -2DR \sin \alpha = L^2-R^2(1-2 \cos \alpha + \cos^2 \alpha)

    The L^2 terms cancel, and if we move the R^2cos^2 term to the left and combine the sine terms we get:

     R^2 (\sin^2 \alpha + \cos^2 \alpha) +D^2 -2LD + 2R(L-D) \sin \alpha = -R^2 +2R^2 \cos \alpha

      = 2R^2 (1- \cos \alpha) +D^2-2LD +2R(L-D) \sin \alpha = 0

    Divide through by 2R^2, and rearrange:

    1+\frac {D^2}{2R^2}-\frac {LD}{R^2} = \cos \alpha - (\frac {L-D} R) \sin \alpha

    Here is where we set  c = \frac {L-D} R and  k = 1+\frac {D^2}{2R^2}-\frac {LD}{R^2}, and replace the cosine with  \cos \alpha = \sqrt{1- \sin^2 \alpha}:

     \sqrt{1- \sin^2 \alpha} = k + c \sin \alpha

    Square both sides:

    1 - \sin^2 \alpha = k^2 + 2kc \sin \alpha + c^2 \sin^2 \alpha

    0 = (c^2+1) \sin^2 \alpha + 2kc \sin \alpha + (k^2-1)

    This is a quadratic in sin alpha. Use the quadratic formula to find two values of sin alpha. Take the arc sine to find the angles. When you do this you should get:

    \alpha_1 = 47.4^{\circ}, \ \alpha_2 = 10.5^{\circ}

    As I noted previously we need to be careful about making sure alpha s in the correct quadrant. That second value for alpha should actually be 180-10.5 = 169.5.
    Last edited by ebaines; July 23rd 2014 at 12:07 PM.
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    Re: Calculate crank angle given pushrod displacement

    Big thanks again for working this through and walking me through the route to it.

    Quote Originally Posted by ebaines View Post

    As I noted previously we need to be careful about making sure alpha s in the correct quadrant. That second value for alpha should actually be 180-10.5 = 169.5.
    Just one final question, without visualising (sketching out) the solutions, is the anyway to deduce the correct quadrants for both solutions? I know sine is positive in the first and second quadrant, but just wonder how you figured the second solution 10.5 and not the first 47.4 needed subtracting from 180 without sketching?
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  15. #15
    MHF Contributor ebaines's Avatar
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    Re: Calculate crank angle given pushrod displacement

    When evaluating an arcsine there is no "automatic" way to know which is the correct quadrant - you have to consider the geometry and see which solution fits. In this case you could calculate the length of displacement D for each of the possible angles: 47.4, 10.5,180-47.4, and 180-10.5 to see which one(s) result in D= 30.
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