This paper Stiffness of System is fourth article about Micro Robot Control in laminar flow thesis. Here is the first one(https://tunavatansever.com/2021/05/09/micro-robot-control/). This time we are considering how to calculate stiffness of the micro robots in laminar flow. Have a nice readings…

In order to understand the stability of our robot, we had to find the stiffness of the robot. Stiffness formula is:

Stiffness:? /∆? (newton\m) (2.10)

We looking for y-axis stiffness. In order to find the value of the stiffness value on the y-axis, we need to calculate the force on the y-axis acting on the robot. The force that affects the robot on the y axis is the drag force. Drag force formula is:  FD =1/2(ρ.V^2.CD) (2.11)

Velocity formula in drag force formula (2.11) is:

? = ? /?

The variables and values in the above equation are given in the table below.

Table 2.3: Quantity of Stiffness Equations

When the data given in the table is written in the equation and the units are equalized,

FD = 1/2( 998,206 . (35/21)^2) . 0,005 =3.85 (Newton) (2.13)

drag force is calculated as 3.85 Newton.

The system has a horizontal length of 0.06 meters. This causes a drop in pressure. This decrease in pressure also causes a reduction in the force acting on the microrobot. The pressure drop mentioned here, the formula for rectangular pipe is:

∆? = ?h? = ??h? (2.14)

h? = 2*? (? *?^2)/ (?h *?) (2.15)

? = 16/Re  (in laminar flow) (2.16)

Re = (v∗?h∗ρ )/ μ (for rectangular duct)  (2.17)

The units and values of the variables in these equations are given in the table below.

Table 2.4: Quantity of Pressure Drop Equation

When the data given in the table is written instead of the equation and the units are equalized pressure drop is calculated as:

∆? = 1.542 (pascal) (2.18)

The change in pressure reduced the force on the y-axis acting on our micro robot.

? = ?/? (2.19)

Calculated from equation (2.19).

? = ?.? (2.20)
? = 1.542*0.09817=0.1513 ??  (2.21)

Stiffness calculated from equation (2.10).

Stiffness: ?/ ∆? = 3.85−0.0000001513/1 ≈3.85 (newton\mm) (2.22)

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