In dc motors, electrical power (Pel) is converted to mechanical power (Pmech). In addition to fri... more In dc motors, electrical power (Pel) is converted to mechanical power (Pmech). In addition to frictional losses, there are power losses in Joules/sec (Iron losses in coreless dc motors are negligible). Pel = Pmech + Pj loss Physically, power is defined as the rate of doing work. For linear motion, power is the product of force multiplied by the distance per unit time. In the case of rotational motion, the analogous calculation for power is the product of torque multiplied by the rotational distance per unit time. Prot = M x ω Where: Prot = rotational mechanical power M = torque ω = angular velocity The most commonly used unit for angular velocity is rev/min (RPM). In calculating rotational power, it is necessary to convert the velocity to units of rad/sec. This is accomplished by simply multiplying the velocity in RPM by the constant (2 x ∏) /60: ωrad = ωrpm x (2∏)/60
In dc motors, electrical power (Pel) is converted to mechanical power (Pmech). In addition to fri... more In dc motors, electrical power (Pel) is converted to mechanical power (Pmech). In addition to frictional losses, there are power losses in Joules/sec (Iron losses in coreless dc motors are negligible). Pel = Pmech + Pj loss Physically, power is defined as the rate of doing work. For linear motion, power is the product of force multiplied by the distance per unit time. In the case of rotational motion, the analogous calculation for power is the product of torque multiplied by the rotational distance per unit time. Prot = M x ω Where: Prot = rotational mechanical power M = torque ω = angular velocity The most commonly used unit for angular velocity is rev/min (RPM). In calculating rotational power, it is necessary to convert the velocity to units of rad/sec. This is accomplished by simply multiplying the velocity in RPM by the constant (2 x ∏) /60: ωrad = ωrpm x (2∏)/60
When selecting drive wheel motors for mobile vehicles, a number of factors must be taken into acc... more When selecting drive wheel motors for mobile vehicles, a number of factors must be taken into account to determine the maximum torque required. The following example presents one method of computing this torque. Example vehicle design criteria: ▪ Gross vehicle weight (GVW): 35 lb ▪ Weight on each drive wheel (WW): 10 lb ▪ Radius of wheel/tire (Rw): 4 in ▪ Desired top speed (Vmax): 1.5 ft/sec ▪ Desired acceleration time (ta): 1 sec ▪ Maximum incline angle (α): 2 degree ▪ Worst working surface: concrete (good) To choose motors capable of producing enough torque to propel the example vehicle, it is necessary to determine the total tractive effort (TTE) requirement for the vehicle: TTE [lb] = RR [lb] + GR [lb] + FA [lb] Where: TTE = total tractive effort [lb] RR = force necessary to overcome rolling resistance [lb] GR = force required to climb a grade [lb] FA = force required to accelerate to final velocity [lb] The components of this equation will be determined in the following steps. Step One: Determine Rolling Resistance Rolling Resistance (RR) is the force necessary to propel a vehicle over a particular surface. The worst possible surface type to be encountered by the vehicle should be factored into the equation. RR [lb] = GVW [lb] x C rr [-] where: RR = rolling resistance [lb] GVW = gross vehicle weight [lb] Crr = surface friction (value from Table 1) Example: RR = 35 lb x 0.01 (" good concrete ") = 0.35 lb Step Two: Determine Grade Resistance Grade Resistance (GR) is the amount of force necessary to move a vehicle up a slope or " grade ". This calculation must be made using the maximum angle or grade the vehicle will be expected to climb in normal operation. To convert incline angle, α, to grade resistance: GR [lb] = GVW [lb] x sin(α) where: GR = grade resistance [lb] GVW = gross vehicle weight [lb] α = maximum incline angle [degrees] Example: GR = 35 lb x sin(2°) = 1.2 lb Step Three: Determine Acceleration Force Acceleration Force (FA) is the force necessary to accelerate from a stop to maximum speed in a desired time. FA [lb] = GVW [lb] x Vmax [ft/s] / (32.2 [ft/s 2 ] x ta [s]) where: FA = acceleration force [lb] GVW = gross vehicle weight [lb] Vmax = maximum speed [ft/s] ta = time required to achieve maximum speed [s] Example: FA = 35 lb x 1.5 ft/s / (32.2 ft/s 2 x 1 s) = 1.6 lb
–Automation systems are generally made up of three main subsystems, namely mechanical, electrical... more –Automation systems are generally made up of three main subsystems, namely mechanical, electrical and software. The interactions among these components affect the integrated system in terms of reliability, quality, scalability, and cost. Therefore, it is imperative that the three components of automation systems are designed concurrently through an integrated design paradigm. This leads to the need to teach integrated design concepts to students in programs such as process automation, electrical and computer engineering, and mechanical engineering. However, due to the time constraint, it is almost impossible to run full integrated design class projects. Therefore, instructors have to decide on the parts of the design process that their class projects have to focus on, and the parts that have to be reviewed for the completeness of the integrated design process. In this paper we present the design and implementation of a microcontroller based, 3D printable, low cost robotic arm suitable for teaching integrated design. Moreover, the paper presents how the robotic arm design is used in an integrated design project of an Industrial Networks and Controllers course. Since the focus of this course is the electrical and software subsystems of the robotic arm, and we do not have enough time to do a full design, students review the design of the robotic arm presented in this paper and use it to either 3D print the robotic arm or purchase the mechanical subsystem of the robotic arm that meets the specification.
In dc motors, electrical power (Pel) is converted to mechanical power (Pmech). In addition to fri... more In dc motors, electrical power (Pel) is converted to mechanical power (Pmech). In addition to frictional losses, there are power losses in Joules/sec (Iron losses in coreless dc motors are negligible). Pel = Pmech + Pj loss Physically, power is defined as the rate of doing work. For linear motion, power is the product of force multiplied by the distance per unit time. In the case of rotational motion, the analogous calculation for power is the product of torque multiplied by the rotational distance per unit time. Prot = M x ω Where: Prot = rotational mechanical power M = torque ω = angular velocity The most commonly used unit for angular velocity is rev/min (RPM). In calculating rotational power, it is necessary to convert the velocity to units of rad/sec. This is accomplished by simply multiplying the velocity in RPM by the constant (2 x ∏) /60: ωrad = ωrpm x (2∏)/60
In dc motors, electrical power (Pel) is converted to mechanical power (Pmech). In addition to fri... more In dc motors, electrical power (Pel) is converted to mechanical power (Pmech). In addition to frictional losses, there are power losses in Joules/sec (Iron losses in coreless dc motors are negligible). Pel = Pmech + Pj loss Physically, power is defined as the rate of doing work. For linear motion, power is the product of force multiplied by the distance per unit time. In the case of rotational motion, the analogous calculation for power is the product of torque multiplied by the rotational distance per unit time. Prot = M x ω Where: Prot = rotational mechanical power M = torque ω = angular velocity The most commonly used unit for angular velocity is rev/min (RPM). In calculating rotational power, it is necessary to convert the velocity to units of rad/sec. This is accomplished by simply multiplying the velocity in RPM by the constant (2 x ∏) /60: ωrad = ωrpm x (2∏)/60
When selecting drive wheel motors for mobile vehicles, a number of factors must be taken into acc... more When selecting drive wheel motors for mobile vehicles, a number of factors must be taken into account to determine the maximum torque required. The following example presents one method of computing this torque. Example vehicle design criteria: ▪ Gross vehicle weight (GVW): 35 lb ▪ Weight on each drive wheel (WW): 10 lb ▪ Radius of wheel/tire (Rw): 4 in ▪ Desired top speed (Vmax): 1.5 ft/sec ▪ Desired acceleration time (ta): 1 sec ▪ Maximum incline angle (α): 2 degree ▪ Worst working surface: concrete (good) To choose motors capable of producing enough torque to propel the example vehicle, it is necessary to determine the total tractive effort (TTE) requirement for the vehicle: TTE [lb] = RR [lb] + GR [lb] + FA [lb] Where: TTE = total tractive effort [lb] RR = force necessary to overcome rolling resistance [lb] GR = force required to climb a grade [lb] FA = force required to accelerate to final velocity [lb] The components of this equation will be determined in the following steps. Step One: Determine Rolling Resistance Rolling Resistance (RR) is the force necessary to propel a vehicle over a particular surface. The worst possible surface type to be encountered by the vehicle should be factored into the equation. RR [lb] = GVW [lb] x C rr [-] where: RR = rolling resistance [lb] GVW = gross vehicle weight [lb] Crr = surface friction (value from Table 1) Example: RR = 35 lb x 0.01 (" good concrete ") = 0.35 lb Step Two: Determine Grade Resistance Grade Resistance (GR) is the amount of force necessary to move a vehicle up a slope or " grade ". This calculation must be made using the maximum angle or grade the vehicle will be expected to climb in normal operation. To convert incline angle, α, to grade resistance: GR [lb] = GVW [lb] x sin(α) where: GR = grade resistance [lb] GVW = gross vehicle weight [lb] α = maximum incline angle [degrees] Example: GR = 35 lb x sin(2°) = 1.2 lb Step Three: Determine Acceleration Force Acceleration Force (FA) is the force necessary to accelerate from a stop to maximum speed in a desired time. FA [lb] = GVW [lb] x Vmax [ft/s] / (32.2 [ft/s 2 ] x ta [s]) where: FA = acceleration force [lb] GVW = gross vehicle weight [lb] Vmax = maximum speed [ft/s] ta = time required to achieve maximum speed [s] Example: FA = 35 lb x 1.5 ft/s / (32.2 ft/s 2 x 1 s) = 1.6 lb
–Automation systems are generally made up of three main subsystems, namely mechanical, electrical... more –Automation systems are generally made up of three main subsystems, namely mechanical, electrical and software. The interactions among these components affect the integrated system in terms of reliability, quality, scalability, and cost. Therefore, it is imperative that the three components of automation systems are designed concurrently through an integrated design paradigm. This leads to the need to teach integrated design concepts to students in programs such as process automation, electrical and computer engineering, and mechanical engineering. However, due to the time constraint, it is almost impossible to run full integrated design class projects. Therefore, instructors have to decide on the parts of the design process that their class projects have to focus on, and the parts that have to be reviewed for the completeness of the integrated design process. In this paper we present the design and implementation of a microcontroller based, 3D printable, low cost robotic arm suitable for teaching integrated design. Moreover, the paper presents how the robotic arm design is used in an integrated design project of an Industrial Networks and Controllers course. Since the focus of this course is the electrical and software subsystems of the robotic arm, and we do not have enough time to do a full design, students review the design of the robotic arm presented in this paper and use it to either 3D print the robotic arm or purchase the mechanical subsystem of the robotic arm that meets the specification.
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