Math and Science Calculations

Rov Arm Calculations
The calculations used in determining the correct components for the hydraulic system are shown below. I needed a hydraulic system that would divide the force at the arm, as only a small force is needed compared to the usual forces used in hydraulic systems. Most hydraulic systems are used to gain a mechanical advantage, such as raising a car with a minimal amount of force. In the case of the Rov, a force of only 2 pounds is needed to clamp onto the heaviest object, and to safely secure it for a trip to the surface a force of 3 pounds would suffice. By examining Pascal’s principle I can prove that my design will function.

Force Required To Operate Arm:
One of the major concerns addressed in my presentations was a kink in the hydraulic line on my model. While a complete kink will cut off flow and make it difficult for the arm to function, minor bends and very long hydraulic lines have no effect. Pascals Equation P1=P2 proves that in a closed system the pressure at any point is equal to the pressure at all other points. Therefore by applying pressure at the control station, the pressure will be transmitted with full efficiency to the Rov arm. Since pressure equals force per unit area the following equation also holds true; F1/A1=F2/A2.
The master piston has a surface area of 1.227 square inches found by using the equation pi x r squared, or ((pi) x (.625)(.625)). The slave piston, found on the Rov, has a surface area of .785 square inches determined using the same formula, or ((pi) x (.5)(.5)). Since a force of 3 pounds is needed at the slave piston the previous equation can be used to determine the force required at the master piston.
F1/A1=F2/A2
(F1)/(1.227)=(3)/(.785)
(F1)/(1.227)=(3.8216)
(F1)=(1.227)x(3.8216)
F1= 4.6891

This equation proves that to deliver a force of 3 pounds at the Rov arm, a force of 4.6891 pounds must be applied at the control station. The airline tubing, which has the lowest maximum pressure rating of all the components, is capable of holding 22 psi. Therefore the system is capable of withstanding the 4.6891 pound force, and is capable of much more.

Distance Pistons Must Move:

Pascals Principle also states that that V1=V2, which means that the volume of fluid displaced on one side is equal to the volume of liquid that will appear on the other side. Since volume equals area x distance the following equation can be substituted. A1/D1=A2/D2. Since the areas have already been calculated, I can determine the distance we must depress the master piston to close the Rov arm. According to the current designs the arm will have to move a maximum of 1.8 inches to move the arm from a fully open position to a fully closed position.
A1=1.227” squared(master piston)
A2=.785” squared (slave piston)
D2= 1.8” ( distance moved by slave piston)
D1= ? (Distance the master piston is required to move)

A1/D1=A2/D2
(1.227) x (D1)=(.785) x (1.8)
(1.227) x (D1)=(1.413)
(D1)=(1.413)/(1.227)
(D1)=(1.1516)
D1= 1.1516”

This equation proves that to fully close the arm which requires the slave piston to extend 1.8 inches, seen in Figure 2, the master piston must be depressed 1.1516 inches, the same holds true when extracting the master piston to reopen the arm.
Figure 2 (Hydraulic Arm)

Mechanical Advantage:
By rearranging the previous equation, the mechanical advantage lost or gained by using a larger piston at the control station than the piston on the arm can be found. This equation is D1/D2=A2/A1=IMA.

D1=1.1516”
D2=1.8”
A2=.785” squared
A1=1.227” squared
IMA=? (Mechanical Disadvantage)

D1/D2=A2/A1=IMA
1.1516/1.8 = .785/1.227 =IMA
.6397=.6397 = IMA

This means that the mechanical advantage is .6397, since this number is less than one there is a disadvantage to using this system. This however was done purposefully to increase the closing speed of the robotic arm.