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Mechanical power transmission

  1. Nov 24, 2009 #1
    Hi all

    Serching along this PF I found this formula to solve the power transmission between piston and crankshaft. I have then drawn this to show it.

    According to this formula we have a maximum (100%) power transmission when angle theta is 90º. I think we should have some power loss even at 90º. Not because of friction or something like that, but because simple geometry.

    Am I wrong? or is there another formula around?

    Thanks a lot in advance.

    http://img195.imageshack.us/img195/3300/pistontocrankshaft.jpg [Broken]
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Nov 25, 2009 #2
    What do you mean by power transmission?

    Also where did you get that equation, because it looks... wrong somehow. I'd like to see how they derived it and that its actually showing.
     
    Last edited: Nov 25, 2009
  4. Nov 25, 2009 #3
    Hi xxChristxx

    Thank you for your prompt response.

    I probably did not use the correct expresion. I was simply referring to the way of solving the crankshaft Torque vs. the force on the piston.

    I got this formula from this thread:

    https://www.physicsforums.com/showthread.php?t=266738

    Which indirectly led me to these two other links:

    http://web.mit.edu/~j_martin/www/pistonphysics.bmp

    https://www.physicsforums.com/attachment.php?attachmentid=16099&d=1225070096

    I think that only in the case of a rod extremely larger than the crank radius we will not have significant power loss when Theta is 90º.

    Is it?

    Thank you in advance.
     
    Last edited by a moderator: Apr 24, 2017
  5. Nov 25, 2009 #4
    Ahhi'm tired and I can't get my derivation to work right, I remember lumping my constants differently so my eq looks slightly different. I've got a bad feeling I've done it right now, but screwed up on the actual peice of work (which is why it looks 'wrong' ir different to me).

    Someone who is acutally good at maths and can derive the eq will help you out.
     
  6. Nov 25, 2009 #5
    No problem, xxChrisxx

    I will try to derive it by myself this evening and I will let you know what I get.

    Thanks anyway.

    .
     
  7. Nov 25, 2009 #6

    Ranger Mike

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    math meets real world..again...you are theoretically correct with your assumption that "According to this formula we have a maximum (100%) power transmission when angle theta is 90º."

    if..if...the force acting on the piston was constant from TDC to 90 degrees of crankshaft rotation..but it is not..at 90 degrees, the piston is one half of the entire stroke..the maximum force applied to the top of the piston occurs at TDC and over a few degrees crank rotation past TDC and diminishes from this max point as various pressure graphs show in another forum..look it up I think it was cylinder pressure of IC engine earlier this week??

    one more huge item that was dismissed in your assumption..you got a whole bunch of parasitic drag caused by piston to skirt clearance, piston drag, piston ring drag , bearing drag and this is huge. Also the rotational and reciprocating weight of the entire power train will impact on total horsepower..
     
    Last edited: Nov 25, 2009
  8. Nov 25, 2009 #7
    .

    Ranger Mike, you are absolutely right, but my question is only focused on the mechanical trasmission losses between piston force F and crankshaft torque T due only to the geometry of the system.

    Finally I have derived this.

    It seems the max eficiency is when theta is arctan l/r.. That would be the crank position in the second image.

    I had no time to derive a straight single formula, so we first get ß and in a second step solve T

    http://img4.imageshack.us/img4/240/pistonandrod20.jpg [Broken]

    Thanks anyway. If you observe something wrong in this formulae, please let me know.

    .
     
    Last edited by a moderator: May 4, 2017
  9. Nov 25, 2009 #8
    If less than 100% of the power goes into turning the crank, it has to go someplace else. Where does it go? heat loss (exhaust gases, friction)? If there is no other place, then 100% does go into turning the crank.
    Bob S
     
  10. Nov 25, 2009 #9
    I think it comes down to how your defining efficiency - if you consider efficiency to be power out / power in (this is usually what is meant by efficiency), then Bob is right and efficiency is not a function of position.

    If you are looking at it from some other point of view (torque, velocity, etc.), then things change, for example, the applied torque results in the maximum force at the piston (along the axis) when theta = 90 (this is also when the piston is at it's maximum velocity, if the crank's angular velocity is constant).

    -Kerry
     
  11. Nov 25, 2009 #10

    Mech_Engineer

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    The equations weren't looking right to me, and it turns out you misapplied the law of sines to find beta. The correct formula should be:

    [tex]\beta=asin\left(\frac{r*sin(\theta)}{l}\right)[/tex]
     
  12. Nov 25, 2009 #11

    Hi, Mech

    By sin-1 I mean asin

    I used r=1 and I used l taking r as the unit in my equation. I will correct it for more clarification.

    Thank you.

    .
     
  13. Nov 25, 2009 #12

    Mech_Engineer

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    Yes, asin and sin-1 are equivalent.

    You never said r=1 anywhere, so you need to leave it in the equation.
     
    Last edited: Nov 25, 2009
  14. Nov 25, 2009 #13
    Thank you, Mech_Engineer, I have already included r in the equation above.

    .
     
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