# Three-position synthesis with specific fixed pivots

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**Adapted from: Kinematics and Dynamics of Machinery by Robert L. Norton.**

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At times the designer would like to design a mechanism that would have to go through three positions. In this case the designer would start with the required three positions of the coupler and would work his way to decide on the location of the two pivot points for the ground link. The location of the two pivot points is not the result of the designer’s choice, but is decided by the geometry of the three positions required. This is not practical; often times the designer is constrained to work with specific pivot points to form the ground link while still achieving the required three positions by the coupler motion. This article discusses one of the simplest methods to go about doing that.

Let’s talk a look at this problem:

Figure1: Problem Statement

Design fourbar linkage to move the link CD shown between the positions C1D1 to C2D2 and then to C3D3. Use specified fixed pivots O2 and O4.

The first thing we need to do is to “invert” the problem. The way to do this is to think about the coupler (C1D1) as the ground instead and the ground (O2O4) as the coupler. For now we have one ground, (C1D1), and we have only one position for the coupler (O2O4). We are going to find two more positions for the now coupler (O2O4) that would correspond to the two other positions (C2D2). To do this we need to define the relation between The coupler and the ground.

Draw construction arcs from point C2 to O2 and from point D2 to O2 whose Radii define the sides of triangle C2O2D2. This defines the relationship of the fixed pivot O2 to the coupler line CD in the **second** coupler position. Draw construction arcs from C2 to O4 and from point D2 to O4 to define the triangle C2O4D2. This defines the relationship of the fixed pivot O4 to the coupler line CD in the **second** position.

After locking the relative location of O2O4 in relation to the second position C2D2, it is time to slide this ground, now coupler, to its new location as if it wasn’t the ground but the coupler. That’s to say, we are trying to answer the question: **if C1D1 is the ground, where would the coupler, O2O4, be in the second position? **

You can imagine this happening by sliding the triangle along the two lines C2C1 and D2D1. This results in a new position for O2’O4’. By doing this, we have pretended that the ground link moved from O2O4 to O2’O4’ instead of the coupler moving from C1D2 to C2D2. We have effectively **inverted** the problem.

Now we have two positions for the now coupler, O2O4; O2O4 and O2’O4’. We need another third position to correspond with the third position C3D3. To do this, we need to repeat the same process again by defining the relation of the link O2O4 in relation to C3D3 after that we need to shift it back to C1D1 as if the link O2O4 was the coupler not the ground.

Performing this will result in the third position O2’’O4’’. This is the relative location of the link O2O4 in its third position as a coupler in relation to the new ground link C1D1. Now we have three positions for the pretended coupler O2O4. O2O4, O2’O4’ and O2’’O4’’

Right now the problem is fully redefined; we have three positions for the newly decided coupler O2O4 and the link CD in the first location C1D1. We can now rename those three positions O2O4, O2’O4’ and O2’’O4’’ to E1F1, E2F2 and E3F3. We can now deal with the three positions as we would normally solve for three position coupler problem. First we need to draw the lines E1E2 and E2E3. Then we have to draw the perpendicular bisectors of these lines. The bisectors will intersect in point G. We repeat the same process for point F; draw lines F1F2 and F2F3 then draw the perpendicular bisector of these lines. These two bisectors will intersect in point H.

Now we have a fourbar linkage, GEFH, in which the coupler, EF, goes through three positions.

If you recall, EF is not actually our coupler. EF represents our ground O2O4. What we need to do now is to re-invert the problem again to its original form. By switching EF to be the ground link and GH to be the coupler link we reach at this linkage.

Now we have our coupler, link GH, that should be connected back to the first position that we wish to achieve; C1D1. By extending link 3, GH, to reach to C1D1 we form our final mechanism that guides the coupler, link 3 (HGCD) through the three required positions.

The final thing we need to do is to add a driver Dyad to drive the link O4H. This is simply done by treating O4H as rocker in a crank rocker design problem. The rocker, now O4H, has to pass through H2 and H3 to result in the required motion. Afterwards, the final mechanism is checked for toggle positions and that it can reach the required positions smoothly and in orderly fashion.

In conclusion, the designer is not limited to the resulting fixed pivots for the solution of three positions coupler problems. By using the illustrated method, the design can specify the required fixed pivots and the required coupler motion and work his way around to getting those two requirements achieved.

**Adapted from: Kinematics and Dynamics of Machinery by Robert L. Norton.**

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