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Homework Help: Finding the maximum acceleration of a crank.

  1. May 14, 2013 #1
    The title should actually read Finding the acceleration of a piston with respect to a cranks angle

    1. The problem statement, all variables and given/known data
    I have found a formula for calculating the acceleration of a piston, with respect to a cranks angle, however I've also found a couple of online calculators that give this result - my problem is that the results are completely different, and I'm not sure which is correct - if either.

    2. Relevant equations

    I am using the formula found on this page:

    On the right hand side I have the formula for acceleration:

    [itex]a = -rw^2(cos\theta+(cos2\theta/n)[/itex]


    [itex]r[/itex] = crank radius
    [itex]l[/itex] = rod length
    [itex]\theta[/itex] = crank angle
    [itex]w[/itex] = crank angular velocity
    [itex]n[/itex] = [itex]l/r[/itex]

    The values I have got are as follows:

    [itex]r[/itex] = 4 (cm)
    [itex]l[/itex] = 15 (cm)
    [itex]\theta[/itex] = 1 (rad)
    [itex]w[/itex] = 50 (rad/s)
    [itex]n[/itex] = 3.75

    Putting all of this into the formula I should get:

    [itex]A = cos\theta = 0.5403[/itex]
    [itex]B = cos2\theta = -0.4161[/itex]
    [itex]C = -rw^2 = - 4 x (50^2) = -10000 [/itex]
    [itex]C x ( A + ( B / 3.75 ) ) = -4293.4[/itex]

    So, the acceleration of the piston at an angle of 50 rads is -4293.4 cm/s, which equates to -4.293 m/s, which I'm guessing means it is going backwards?

    There are a couple of problems here. It seems after playing with the values within a spreadsheet, the minimum acceleration is approx -12.6666, yet the maximum acceleration is 7.3541, why such a difference, surely the maximum acceleration to be equal in both directions?

    In comparison, an online calculator I found here (http://www.bigboyzcycles.com/PistonSpeed.htm, when given the following data:

    Stroke (inches): [itex] ( r x 2 ) / 2.54 = 3.1496 [/itex]
    Crankshaft Speed (RPMs): [itex] w / 0.104719755 = 477.465 [/itex]
    Connecting rod length (inches): [itex] l / 2.54 = 5.905511 [/itex]

    Gives 416 ft/s, which equates to 126m/s...

    So, for reasons unknown to me I'm getting a huge difference between these calculations. Neither one is giving me much confidence.

    Thank you.
    Last edited: May 15, 2013
  2. jcsd
  3. May 14, 2013 #2


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    First of all, Please state the complete problem exactly as it was given to you.

    Your title suggests finding the acceleration of the crank , not the piston.
  4. May 15, 2013 #3


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    -4293.4 cm/s = -42.934 meters / s (Remember, there are 100 cm in 1 meter)
  5. May 15, 2013 #4
    Hello SammyS,

    Unfortunately, I cannot as this isn't homework, it's for a personal project I'm working on, and I'm afraid I'm not anything like a physics student so my knowledge may have some very large gaps in it. :) I've clarified the title and the wording in a couple of places. I'm after the acceleration of the piston given a particular crank angle.

    Hello SteamKing,
    Oh! That was a silly mistake I made! Thank you for that, however the numbers are still wildly different... :S

    Thank you both.
  6. May 15, 2013 #5
    It looks like the acceleration varies with crank angle. Putting θ = 1 (rad) into the formula gives the acceleration at that particular position, not the maximum value.

    I get 113.96 m/s^2 as the maximum acceleration from your formula. I think that's close but not very accurate, because that formula is just a approximation.
    Last edited: May 15, 2013
  7. May 15, 2013 #6
    Hello davidchen,

    Thanks for taking the time to reply.

    Yes, that's correct and is what I'm looking for - my apologies, I messed up the title...

    Hm, well as you've noticed the formula gives the acceleration of the piston at a given crank angle. I'm not knowledgeable enough with maths to be able to work out how to get the maximum acceleration - but I'm able to keep adjusting the angle within an excel worksheet to get the maximum. The maximum acceleration I get is 73.54146 and the minimum acceleration (I'm guessing this means it is going backwards) is -126.6668.

    Why should the maximum be different from the minimum? I would have expected an acceleration value of ±126.6668.

    Also, it seems your maximum acceleration is 113.96 whereas mine is 73.54146? However, my minimum acceleration (-126.6668) is closer to your maximum?

    Thanks for your help.
  8. May 15, 2013 #7
    Sorry, I seemed to forgot that minus sign and messed up something. Your calculation is correct. [tex]cos\theta +(cos2\theta /n)[/tex] has a range from -353/480 to 19/15, which is shown in the link.http://www.wolframalpha.com/input/?i=cos(x)+cos(2x)/3.75&t=crmtb01 Put that back into the equation gives your results.
  9. May 15, 2013 #8
    Thank for clarifying that - well, it would seem that the formula is working, however from a practical point of view, it makes little sense.

    If you assume a crank as depicted from wiki:
    http://upload.wikimedia.org/wikipedia/commons/8/8b/Crank_mechanism_geometry_sk.png then why would the acceleration of the piston be greater (i.e. moving 'up' the screen) than when the the crank is pulling the piston (moving 'down' the screen)?

    This seems I may have been working with the wrong formula, or the formula is incorrect for the expected results?

    Thank you.
  10. May 15, 2013 #9
    Consider a small angle (delta theta) at the top of circle, and another one at the bottom. The rod length and angular velocity are constant. So moving through those two sections (two small angles at top and bottom) takes same time. But how much distance does the crank move? Looks different. So this case is not symmetric.
  11. May 15, 2013 #10

    Very sorry - ignore this post.
  12. May 15, 2013 #11
    Hi David. That's excellent - thank you very much for all your help and patience through this thread - it's greatly appreciated. Even though it was a straight forward case of putting values into a formula, I struggled quite a bit with this one. :)

    Many thanks once more to SteamKing as well.
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