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Thread: Practical problem with trignometric functions: crank, connecting rod of steam eng.

  1. #1
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    Practical problem with trignometric functions: crank, connecting rod of steam eng.

    I am reading a book on calculus and I got to the part about trigonometry. There is an illustrative problem for the functions.
    " The crank and connecting rod of a steam engine are three and ten feet long respectively, and the crank revolves at a uniform rate of 120 r.p.m. At what rate is the crosshead moving when the crank makes an angle of 45 degrees with the dead center line?"

    Untitled.jpg


    Now before I read the solution in the book I tried to solve it myself. I expressed the relation by which the crosshead moves as
    x = r - r cos(angle) (in the picture what i call x is the distance from point A to the circle, they call x the distance from O to C),
    because the crosshead is x away from the furthest right point it can go. At 90deg it is exactly r away from that point, at 180deg 2r and then back. I then differentiated this equation and I got
    dx = d(r) - d(rcos(angle)),
    dx = 0 - r d(cos(angle)),
    dx = r sin(angle) d(angle),
    dx/dt = 3 sqrt(2)/2 4pi/sec,
    dx/dt = 26.65ft/sec

    However this wasn't the right solution. In the book they give the solution that involves expressing the length from the center of the crank to the rod. So they then divide the resulting triangle into 2 right-angle triangles (OPA and PAC). Then x = OA + AC; x = r cos(angle) + sqrt(a2 sin2(angle) + b2). When this is differentiated to the end they get(they call radius a):
    dx/dt=-a sin(angle)[1+ (cos(angle)/(sqrt(b/a)2- sin2(angle))]d(angle)/dt

    Now when you insert the known quantities you get dx/dt = -32.44ft/sec(minus because it's moving left,mine was the opposite).
    I don't see how the solutions can be different, or why the amount by which the crank moves at all depends on b. Shouldn't it only depend on the radius since the crank can't move more to the left or right than the raidus)?


  2. #2
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    Quote Originally Posted by skugz View Post
    I am reading a book on calculus and I got to the part about trigonometry. There is an illustrative problem for the functions.
    " The crank and connecting rod of a steam engine are three and ten feet long respectively, and the crank revolves at a uniform rate of 120 r.p.m. At what rate is the crosshead moving when the crank makes an angle of 45 degrees with the dead center line?"

    Untitled.jpg


    Now before I read the solution in the book I tried to solve it myself. I expressed the relation by which the crosshead moves as
    x = r - r cos(angle) (in the picture what i call x is the distance from point A to the circle, they call x the distance from O to C),
    because the crosshead is x away from the furthest right point it can go. At 90deg it is exactly r away from that point, at 180deg 2r and then back. I then differentiated this equation and I got
    dx = d(r) - d(rcos(angle)),
    dx = 0 - r d(cos(angle)),
    dx = r sin(angle) d(angle),
    dx/dt = 3 sqrt(2)/2 4pi/sec,
    dx/dt = 26.65ft/sec
    .....................................But that is not the speed of C (which is what you need to calculate and the book calculated).

    If I define your 'x' as xs and the book's 'x' as xb, then:

    xb = AC - xs + r

    xb = √[b2 - r2sin2(Θ)] - xs + r


    Thus xb and xs are related through Θ and their are rate of change are not equal.
    However this wasn't the right solution. In the book they give the solution that involves expressing the length from the center of the crank to the rod. So they then divide the resulting triangle into 2 right-angle triangles (OPA and PAC). Then x = OA + AC; x = r cos(angle) + sqrt(a2 sin2(angle) + b2). When this is differentiated to the end they get(they call radius a):
    dx/dt=-a sin(angle)[1+ (cos(angle)/(sqrt(b/a)2- sin2(angle))]d(angle)/dt

    Now when you insert the known quantities you get dx/dt = -32.44ft/sec(minus because it's moving left,mine was the opposite).
    I don't see how the solutions can be different, or why the amount by which the crank moves at all depends on b. Shouldn't it only depend on the radius since the crank can't move more to the left or right than the radius)?
    why the amount by which the crank moves at all depends on b?

    The size of 'b' defines the angular motion of CP - which in turn is a factor of linear speed of C.
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

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    If I'm understanding correctly the crank then moves at different speeds then what I call xs but the speeds are equal at times when the angle is 0, 90, 180, 270.

    I still have to get a more intuitive feeling about it but thanks for your help.

  4. #4
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    connecting rod intuition

    Quote Originally Posted by skugz View Post
    I am reading a book on calculus and I got to the part about trigonometry. There is an illustrative problem for the functions.
    " The crank and connecting rod of a steam engine are three and ten feet long respectively, and the crank revolves at a uniform rate of 120 r.p.m. At what rate is the crosshead moving when the crank makes an angle of 45 degrees with the dead center line?"

    Untitled.jpg


    Now before I read the solution in the book I tried to solve it myself. I expressed the relation by which the crosshead moves as
    x = r - r cos(angle) (in the picture what i call x is the distance from point A to the circle, they call x the distance from O to C),
    because the crosshead is x away from the furthest right point it can go. At 90deg it is exactly r away from that point, at 180deg 2r and then back. I then differentiated this equation and I got
    dx = d(r) - d(rcos(angle)),
    dx = 0 - r d(cos(angle)),
    dx = r sin(angle) d(angle),
    dx/dt = 3 sqrt(2)/2 4pi/sec,
    dx/dt = 26.65ft/sec

    However this wasn't the right solution. In the book they give the solution that involves expressing the length from the center of the crank to the rod. So they then divide the resulting triangle into 2 right-angle triangles (OPA and PAC). Then x = OA + AC; x = r cos(angle) + sqrt(a2 sin2(angle) + b2). When this is differentiated to the end they get(they call radius a):
    dx/dt=-a sin(angle)[1+ (cos(angle)/(sqrt(b/a)2- sin2(angle))]d(angle)/dt

    Now when you insert the known quantities you get dx/dt = -32.44ft/sec(minus because it's moving left,mine was the opposite).
    I don't see how the solutions can be different, or why the amount by which the crank moves at all depends on b. Shouldn't it only depend on the radius since the crank can't move more to the left or right than the raidus)?
    intuitively, it does seem pt C would move to the left the same as point P, and length b would not matter.
    however, point P also has a vertical component to its motion.
    imagine pt P moving (straight) upward only. It then drags pt C to the left.
    that is the movement that is added to pt P's horizontal motion, to give you motion of C.

  5. #5
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    draw it

    Quote Originally Posted by skugz View Post
    If I'm understanding correctly the crank then moves at different speeds then what I call xs but the speeds are equal at times when the angle is 0, 90, 180, 270.

    I still have to get a more intuitive feeling about it but thanks for your help.
    the crank rotates at constant speed, but the slider moves at a different speed depending on the angle of the crank. [slider speed is 0 at 00 and 1800, max speed at 900 and 2700]

    to get an intuitive feel- draw the figure to scale and solve it with geometry.
    draw the direction of point P (tangent to the circle) with its speed to a scale of your choice.
    then Px will be its horizontal component, Py its vertical.
    Px=Psin(angle), Py=Pcos(angle) [sin and cos are reversed at point P. you will see this when you draw it]
    then you can see both components move point C, and you can calculate C geometrically.
    you will get the same answer. [in fact, you will derive an equation that provides the same results; it just may look some different.]

    I encourage you to draw it to scale, or at least draw it out and give it some thought.
    it will come to you.
    Last edited by sinx; 03-09-2018 at 06:11 PM.

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    correction to mistake on the post "draw it"

    Quote Originally Posted by skugz View Post
    If I'm understanding correctly the crank then moves at different speeds then what I call xs but the speeds are equal at times when the angle is 0, 90, 180, 270.

    I still have to get a more intuitive feeling about it but thanks for your help.
    correction to mistake on the post "draw it"
    for drawing the components of P; I meant Px=Psin45, and Py=Pcos45.
    [on my post, i had both Px=Py​=Pcos45. They are equal, but only because the angle is 45 degrees.]
    Last edited by sinx; 03-09-2018 at 06:15 PM.

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    more on intuition-

    Quote Originally Posted by skugz View Post
    If I'm understanding correctly the crank then moves at different speeds then what I call xs but the speeds are equal at times when the angle is 0, 90, 180, 270.

    I still have to get a more intuitive feeling about it but thanks for your help.
    more on intuition-
    the difference between 32.44 and 26.65 [=5.784] is the horizontal speed of pt C imposed by the vertical speed of pt. P.
    5.784 = 26.65 tan (phi), where phi is the angle between b and x.
    [26.54 is the vertical speed of P, (the same as horizontal speed, since the angle is 45).]

    to see that tan (phi) is correct, draw the picture and focus on the motion of P on link b.
    as P moves up, and b rotates about C, how much is P (and C) forced to move horizontally?

    then the "geometric" soln for the total speed of C, (for a crank angle of 45)=
    = [a sin45 + a cos45 tan(phi)] [angular velocity of a].
    = [dx/d(angle)] [d(angle)/dt] ; which= dx/dt you found earlier.

    note: The only variables in the "geometric" soln are crank angle and phi; and phi can be expressed in terms of the crank angle (and constants). If you do this, you will get the identical eqn for dx/dt, in one variable (angle).
    Last edited by sinx; 03-09-2018 at 06:22 PM.

  8. #8
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    Quote Originally Posted by skugz View Post
    At 90deg it is exactly r away from that point, at 180deg 2r and then back.
    At 90 degrees it's not r away from that point. It would be, if the rod b was attached to O. But since it's attached to P, it's closer to O. Try drawing the 90 degree case. It does not represent the middle point between the 2 extremes.

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