The OP's second link, gives damping rates albeit in a most confusing way. If you work it out, for example, the first damper has 1.4 in/sec as the damping and references 40 lbf which means that the damper applies 40lbf when a velocity of 1.4 in/sec is applied. In this case: $$C_{example} = \frac{40(lbf)}{1.4 \frac{in}{sec}} = 28.6(\frac{lbf\cdot sec}{ft})$$.

Which you multiply by the velocity to get the force.

$$F_{damping}(lbf) = V (\frac{ft}{s})*C(\frac{lbf\cdot sec}{ft})$$

The OP's second link, gives damping rates albeit in a most confusing way. If you work it out, for example, the first damper has 1.4 in/sec as the damping and references 40 lbf which means that the damper applies 40lbf when a velocity of 1.4 in/sec is applied. In this case: $$C_{example} = \frac{40(lbf)}{1.4 \frac{in}{sec}} = 343(\frac{lbf\cdot sec}{ft})$$

Which you multiply by the velocity to get the force:

$$F_{damping}(lbf) = V (\frac{ft}{s})*C(\frac{lbf\cdot sec}{ft})$$

The OP's second link, gives damping rates albeit in a most confusing way. If you work it out, for example, the first damper has 1.4 in/sec as the damping and references 40 lbf which means that the damper applies 40lbf when a velocity of 1.4 in/sec is applied. That equates to damping of 28.6 (lbf*s)/ft.

The OP's second link, gives damping rates albeit in a most confusing way. If you work it out, for example, the first damper has 1.4 in/sec as the damping and references 40 lbf which means that the damper applies 40lbf when a velocity of 1.4 in/sec is applied. In this case: $$C_{example} = \frac{40(lbf)}{1.4 \frac{in}{sec}} = 28.6(\frac{lbf\cdot sec}{ft})$$.

Which you multiply by the velocity to get the force.

$$F_{damping}(lbf) = V (\frac{ft}{s})*C(\frac{lbf\cdot sec}{ft})$$