Fourbar Linkages

A fourbar linkage consists of four links and four joints, in which one link is usually grounded (fixed).

If S is the length of the shortest link, L is the length of the longest link, and P and Q are the lengths of the remaining two links, then the formula below shows the Grashof condition. If the Grashof condition is true for a given fourbar linkage, then the mechanism will be able to perform a full rotation with respect to the ground.

S + L <= P + Q

Fourbar Linkage Analysis

Fourbar mechanisms can be analyzed to find the positions, velocities, and accelerations of different parts based on the angular rotation of each joint. The image below shows a fourbar linkage with angles from the positive x-axis to the link.

If we assume that theta 1 is at the origin, then by following the vectors we can find the following equation:

L₁ + L₂ - L₃ - L₄ = 0

because by following the shape in the direction of L₁ and L₂ and then following vectors L₃ and L₄ backwards, you arrive at the origin again, causing 0 displacement.

Each link can be represented by the following:

L_a = l_ae^jӨ

in which Ө represents the link's angle from the positive x-axis and l_a represents the length of L_a. So if we plug that in to the above equation:

l₁e^jӨ₁ + l₂e^jӨ₂ - l₃e^jӨ₄ - l₄e^j0 = 0

Notice that l₄ has an angle of zero since it is parallel to the positive x-axis.

Using Euler's formula, e^jӨ = cos(Ө) + jsin(Ө), we can separate this equation into two equations, one using the real values and one using the imaginary by dividing all terms by j:

Real: l₁cos(Ө₁) + l₂cos(Ө₂) - l₃cos(Ө₄) - l_{4 = 0}

Imaginary: l₁sin(Ө₁) + l₂sin(Ө₂) - l₃sin(Ө₄) = 0

Now we can rearrange each equation:

l₁cos(Ө₁) = l₃cos(Ө₄) + l_{4 - l2cos(Ө2)}

l₁sin(Ө₁) = l₃sin(Ө₄) - l₂sin(Ө₂)

Square them:

l₁²cos²(Ө₁) = l₃²cos²(Ө₄)