Physics Help Forum Whitworth quick return mechanism

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

 Aug 8th 2016, 01:52 AM #1 Junior Member   Join Date: Jul 2016 Posts: 17 Whitworth quick return mechanism See the picture here: https://pl.vc/zk3gn The transmission behavior defined by the distance a between 2 bearing points and the length b>a of the front-end crank which rotates at constant angular velocity, i.e. β=ω0.t Express the position B as function of β Express tanα as function of β Compute angular velocity ω=α˙(t)of output crank! (I meant it's the derivative of alpha) I've tried to solve 1st question as: x = a + OA cosβ y = OA sinβ But I have no idea how to answer the 2nd and 3rd question. From the diagram, the relation that I can make is: Triangle ABO: (OA)^2=a^2 + (BA)^2 − 2 a OB cosα Triangle AOC (I made my own C): (OA sinβ)^2 = (OA)^2+ (OA cosβ)^2 − 2 (OA) (AB cosα) cosα Triangle ABC: (BA sinα)^2 = (a + OA sinβ)^2 + (BA)^2 − 2(BA)(a + OA cosβ) cosα Is it right if I solve tan α with the value of sinα from triangle ABC and the value of cos α from triangle ABO? Last edited by rndaryam; Aug 8th 2016 at 02:02 AM.
 Aug 8th 2016, 05:43 AM #2 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,324 It seems you have defined (x,y) as the position of A, but the question asked about the position of B. You need to clarify how the mechanism is constructed. The drawing makes it look like the points A, B and C are all pivots, but that would mean the mechanism can't turn. Perhaps as crank OA rotates the tip of the AB linkage at B slides along the y-axis, so that the length 'a' varies? But if so, then it makes no sense to ask about the value of y for point B as it is always zero. So clearly I don't understand the mechanism. Also, is the length 'b' the same as the length OA? Last edited by ChipB; Aug 8th 2016 at 08:30 AM.
 Aug 8th 2016, 08:46 AM #3 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,324 OK, I think I see what you are doing. Points O and B are fixed pivots, and point A slides along the linkage that pivots at B, as in the attached sketch, right? In other words length OB=a and OA=b are fixed, and the length AB varies as the crank OA turns. If so, the value of x and y that you calculated are the coordinate for point A, not B. So then tan(alpha) = y/x = [b sin(beta)]/[a + b cos(beta)]. Then determine the derivative of both sides with respect to time, and solve for d(alpha)/dt. Attached Thumbnails   Last edited by ChipB; Aug 8th 2016 at 12:43 PM.
 Aug 9th 2016, 04:01 PM #4 Senior Member   Join Date: Jun 2010 Location: NC Posts: 405 I don't know why you are working this or what level you are. There is a powerful method available. The plan is to work in two dimensions with vectors. The mechanism is "constrained." Start with a vector description of the 2-D situation. All else is simply paying attention. I'm cooking tonight! I wrote some of the equations of your case. See it here. http://www.thermospokenhere.com/wp/0...ck_return.html What is the derivative of a time-dependent unit vector? http://www.thermospokenhere.com/wp/01_tsh/A9820_1.25_vectors_circular/vectors_circular.html I'll do more upon request... La ti da, JP

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