EMERSON XVM-403-TBNS-0000 motor

Specify the master interval and starting point. Next we need to construct the ECAM table. The table is specified at uniform intervals of master positions. Up to 256 intervals are allowed. The size of the master interval and the starting point are specified by the instruction:where m is the interval width in counts, and n is the starting point. For the given example, we can specify the table by specifying the position at the master points of 0, 2000, 4000 and 6000. We can specify that by EP 2000,0 Step 4. Specify the slave positions. Next, we specify the slave positions with the instruction ET[n]=x,y,z,w where n indicates the order of the point. The value, n, starts at zero and may go up to 256. The parameters x,y,z,w indicate the corresponding slave position. For this example, the table may be specified by

This disengages the slave axis at a specified master position. 

If the parameter is outside the master cycle, the stopping is instantaneous. To illustrate the complete process, consider the cam relationship described by the equation: Y = 0.5 * X + 100 sin (0.18*X) where X is the master, with a cycle of 2000 counts. The cam table can be constructed manually, point by point, or automatically by a program. The following program includes the set-up. The instruction EAX defines X as the master axis. The cycle of the master is 2000. Over that cycle, Y varies by 1000. This leads to the instruction EM 2000,1000.

Suppose we want to define a table with 100 segments.

 This implies increments of 20 counts each. If the master points are to start at zero, the required instruction is EP 20,0. The following routine computes the table points. As the phase equals 0.18X and X varies in increments of 20, the phase varies by increments of 3.6°. The program then computes the values of Y according to the equation and assigns the values to the table with the instruction ET[N] = ,Y.Now suppose that the slave axis is engaged with a start signal, input 1, but that both the engagement and disengagement points must be done at the center of the cycle: X = 1000 and Y = 500. This implies that Y must be driven to that point to avoid a jump

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