Dynamic Aspects Relating to the Design of a Quasi-motion Platform Douglas Gobeski 11/07/08 Abstract To give a higher-quality simulator experience, a platform is constructed for the user to stand on. The platform rotates through a limited distance about two axes to give the sensation of motion. The actuators that are used to tip the platform up and down have a finite ability to generate force, and therefore must be designed to provide sufficient force, angular displacement, and angular frequency. A distance of eighteen inches between the pivot point and each actuator was chosen to satisfy these competing requirements. Keywords Actuator – a device which only moves up and down Polar Moment of Inertia – resistance to angular acceleration; the higher the polar moment of inertia, the harder it is to speed up/slow down rotation Quasi-motion – Rotational motion about two axes only; no actual translational motion, but intended to give the impression of translational motion Introduction In order to give a realistic feeling of motion to the user of a computer simulation, a platform is designed for them to stand on. Through the use of hydraulic actuators, the platform oscillates about a central point through a limited range of motion. This provides the user with a feeling of acceleration without significant platform motion. Thus, themain design concerns from a mechanical standpoint are geometrical in nature and require careful consideration. Objective This application note is intended to provide insight and instruction into the mechanical aspects of the design of a quasi-motion platform for computer simulation use. It will cover the gross geometrical concerns as well as the dynamic considerations. Issues The first issue in designing a quasi-motion platform lies in the determination of the overall geometry of the platform. Although this has significant implications on the performance and weight of the platform, it is primarily a function of the application for which the platform is being designed. For the current application, which is the simulation of the pilot cabin of a tugboat, the dimensions selected were six feet front-to-back and eight feet side-to-side. These dimensions were chosen because they would allow the platform to closely approximate the dimensions of the actual tugboat being simulated. Once the size of the footprint is decided, the spacing of the hydraulic actuators must be determined. This decision lies at the heart of the design of the motion platform. The hydraulic actuators can only produce a limited amount of force. In this application, the hydraulic actuators are Goodyear air cushion springs which have been filled with oil instead of air. Each actuator has a diameter of 4”, and can handle pressures up to 100 psi. As force is equal to the product of pressure and area, the resultant force output is approximately 1250 lbs. ( ).1257.2**100**22lbsFinpsiFrPF=⇒=⇒=ππAs a result of the limit placed on the force that the actuators can produce, a corresponding limit is also placed on the platform dynamics. The reason for this is that the rotational motion of oscillating the platform requires the application of a torque to the platform. Torque ()τ is defined as the product of force and the distance between the rotational center and the applied torque. ()dF *=τ The force is now fixed at 1250 lbs., which sets the torque at 1250*d ft-lbs. The “d” term represents the distance between the pivot point and an actuator. Here, a trade-off exists. As the actuators are spaced out further and further from the pivot point, the torque produced increases linearly. Simultaneously, the maximum amplitude of the oscillation is decreases linearly. It is desirable to maximize the amplitude, while still having sufficient torque to move the platform. Therefore, the minimum necessary torque must be determined. Here, torque becomes ατ*J=, where J is the polar moment of inertia of the platform and α is the angular acceleration of the platform. α is equal to the product of the maximum angular amplitude and the square of the frequency of oscillation. ()2* fA=α The higher the frequency, the faster the platform will move. To estimate the polar moment of inertia of the platform, the masses and positions relative to the center of rotation were estimated for the steering console, ceiling, floor, front wall, side wall (x2), and user (x2). Once each was determined, they were summed together to produce an estimate of the polar moment of inertia (J). The formula used fordetermining the polar moment of inertia for each component was ()()[]( )( )+++2222*12*rotationofcentertocomponentofcenterfromdistxrotationofcentertocomponentofcenterfromdistycomponentofmasscomponentofmasscomponentofwidthcomponentofheight The resultant estimate for polar moment of inertia was approximately 920 slug-ft2, and the estimated total mass was 1300 lbs. Since the total mass is distributed between the actuators and the pivot point (which is a universal joint that acts as a structural component), the static load is not going to exceed 1250 lbs. on the actuator. Putting all of the equations used so far together, the equation becomes 22**736.0 fAftsd = , where A is the angular amplitude of the oscillation, and f is the frequency of the oscillation. Conclusions A set of frequencies from 0.1 Hz to 0.5 Hz were examined in 0.1 Hz intervals, for angular amplitudes of 1° to 15° in 1° increments. (see Table 1) All calculations were done in Microsoft Excel. This analysis showed that with a factor of safety of two, and eight inches separating the top platform and bottom platform at the pivot point, all angular amplitudes could be accommodated at a frequency of 0.3 Hz when the actuators were placed eighteen inches from the pivot point (see Table 2). This would compress one actuator to five inches and expand the opposite actuator to eleven inches. This was deemed acceptable because it would limit the expansion of the actuators to less than 50% of their nominal height. In this case, the decision was made to favor the range of motion over the speed of motion.Table 1. Frequency v. Amplitude alpha Amplitude (deg.) Torque needed (ft-lb) Amplitude (deg.) Torque needed (ft-lb) 0.1 Hz 0.00689 1 6.331012526 0.4 Hz 0.110245 1 101.2962004 0.013781 2 12.66202505 0.220489 2 202.5924008 0.020671 3 18.99303758 0.330734 3 303.8886012 0.027561 4 25.3240501 0.440978
View Full Document
Unlocking...