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Chain pain

  1. Mar 2, 2004 #1
    Hi
    Ive been trying to solve the following problem:
    There are two gear wheels whose centers are seperated by 25 cm.
    a chain is wrapped around the two wheels such that if one wheel is turned the other turns(much like a cycle chain.) If th radius of the wheels are 11cm & 4 cm respectively, find the lenth of the chain.

    I know the problem seems simple, and i have solved it. But I did it by estending the chains in the direction of the smaller wheel and ended up with a cone. Then I foundout the length of thechain,but I realise that the working was too long. Is there a simpler method to solve this . Is there a general method?
    Plz help.
     
    Last edited: Mar 2, 2004
  2. jcsd
  3. Mar 2, 2004 #2

    Integral

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    The general method for this type of problem is to draw a very careful picture then use trig and geometry to arrive at an equation. Here is what I came up with.

    Let:
    D = seperation of the centers of the gears
    S = length of chain between the gears (ie length of the line tangent to both circles)
    [tex] R_1 = {radius\ of\ large\ gear} [/tex]
    [tex] R_2 = {radius\ of\ small\ gear}[/tex]
    [tex] \delta R = R_1 - R_2 [/tex]
    [tex] \theta = sin^{-1}( \frac { \delta R} D)[/tex]
    I have to go, work calls will complete the expression later!
     
    Last edited: Mar 2, 2004
  4. Mar 2, 2004 #3
    could u plz explain further. why did u subtract R2 from R1.
    is
    PHP:
    theta
    the angle subtended by the part of the chain wrapped around the centre.

    Also how do I add a figure to my post. I tried attaching a paint file , but it is always to large.


    THanks in advance.
     
  5. Mar 2, 2004 #4

    Integral

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    Ok, am back!

    The line which is tangent to each of the circles (gears) is perpendicular to a radius of each gear, if we translate this line , maintaining the right angles, down the radius of the small gear you have a right triangle with the small side = to the difference in radius and the hypotenuse the separation of the gears. This is where I get [tex] \theta [/tex] which is the angle the chain makes with the center line. So now we can compute the length of the tangent line as
    [tex]S = D Cos( \theta) [/tex]

    By examining the geometry you will find that [tex] \theta [/tex] is also the "over lap" angle. That is on the larger gear the chain must be in contact with the gear a bit more then Pi radians and on the small gear a bit less then Pi radians, this bit is [tex] \theta [/tex] so for the total length (L) of chain we sum up these pieces.

    [tex] L = 2S + ( \Pi - 2 \theta) R_2 + ( \Pi + 2 \theta)R_1 [/tex]

    EDIT: click on the formulas to see the code that creates them. Also see the LaTex stickies at the top of General Physics.
    EDIT (again!): BTW using this formula I get L=99.1cm for your gears.
     
    Last edited: Mar 2, 2004
  6. Mar 3, 2004 #5
    thanks a million.
     
  7. Mar 11, 2004 #6
    I was wondering if i cud see your diagram... i'm convinced with the final formula. However, i dont fully understand how S = Dcos(theta)... shudn't it b different from Dcos(theta) because the length of chain around each wheel is a little off pie... had the length of the chain around each wheel been pie, then S wud equals Dcos(theta).
    Its hard to explain what i mean without a diagram...
     
  8. Mar 14, 2004 #7

    Integral

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    Here is a drawing I made in ACAD. This is a scale representation of the gears described above. Note that S corresponds to the distance 24 as labeled in the diagram. The key is to understand that [tex] \theta[/tex] is the angle between the vertical and the point at which the chain is tangent to the gear. Since the chain is tangent to both gears it must be perpendicular to the radius at the point of contact on BOTH gears.
     
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