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The accuracy and flexibility of the operator determine the operating efficiency of the robot.
#Degree of freedom free
Since the total number of categorical observations was known in advance, and the other three categorical quantities were also known, then the “lost” categorical quantity can be determined by algebraic solution – its value is “not free to vary.” Hence, the degrees-of-freedom (df) would be expressed as df = c –1 = 4 – 1 = 3, where c is the total number of categories.With the rapid development of mechatronics and robotics technology, the application of robots has been extended from the industrial field to daily life and has become an indispensable part of work and daily life. Solving for D, we note that D = N – (A + B + C), or D = 10 – ( 2 + 3 + 1) = 10 – 6 = 4. Thus, it can be reasoned that N – D = A + B + C. Third, we know that the exact number of observations assigned to the other three categories (A = 2, B = 3, and C = 1). Well, this would not be a problem because of three things.įirst, we know the total number of observations that were made (N = 10). It is known that, of the 10 categorical observations, 2 were assigned to category “A,” 3 were assigned to category “B,” and 1 was assigned to category “C.” However, let us say that we somehow lost all of the observations related to category “D.” In this case, we don’t know how many type “D” observations were made. Now let us postulate that we previously studied the four categories and made a total of 10 observations.
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Say categories A, B, C, and D, respectively. Thus, for the given scenario, we have N – 2 = 6 degrees-of-freedom.Īs a final analogy let us say that there exists four independent categories related to some phenomenon. Given these restrictions, it should now be apparent, that if we grasp any two of the beads, so as to meet the goal, then N –2 = 8 – 2 = 6 of the beads would be “fee to vary” somewhere along the rod, but 2 of the 8 could not vary (as they are used to fix the rod in space). Handpicked Content: How do you determine the Xs in the analyze phase of DMAIC? Only in this manner (by definition), can the “rod of beads” be properly suspended in space. To do this, we will assert that the left-hand must grasp a bead, as well as the right-hand. For the sake of discussion, it will be known that the goal is to suspend or otherwise hold the rod of beads in space (parallel to the ground). Further imagine that this is done in such a way that the beads are “free” to move anywhere on the rod, but without falling off. Now, imagine that we thread each of the 8 beads onto the wire rod. So as to work with some “hard numbers,” it will be known that there is only 1 wire rod and 8 beads, where each bead has a small hole in the middle.
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Of course, we are talking about something like coat hanger wire and costume jewelry beads. Suppose we have before us a simple wire with some beads on it. When considering the last position (and player) there is no choice of selection (or assignment) – simply because the total number of positions and players is fixed in advanced (by virtue of the game’s rules). In other words, one degree-of-freedom (decision option) is lost every time a position-player decision is made. In other words, the coach is not “free” to pick either the last position or the last player.įor every position that is assigned, the remaining number of choices (decision options) is decreased by one. Once the 8th player is assigned to the 8th position, the 9th player-position is pre-determined, so to speak. The coach is “free” to assign any of the 9 players to any of the 9 positions. We understand the field-of-play consists of 9 positions. To facilitate an answer, we will consider three examples.