How can we get floating point resolution using SINT?

The Ethernet tables within the Hardy unit are 16 bit registers. If you select either an INT or FLOAT as the form of data transfer, they will use two registers in the Hardy tables. Example, maps EIO0=+HFI0, and EFO2=+HFI0, both map the gross weight to the Ethernet output table in our unit. The first is an INT and the second is FLOAT. The EIO0 (Ethernet INT output word 0) will use our table words 0 & 1 and EFO2 (Ethernet FLOAT output word 2) will use our table words 2 & 3.

Now realize that when we make a selection of INT or FLOAT that will determine what the format of the data in these registers will be. If INT is selected, then we will transfer the integer value to the left of the decimal point and it will be in integer format. If FLOAT is selected, then we will transfer the entire value in IEEE float format. Both these values, since they are 32 bit values, will be "cut" in half and so you will get the least significant 16 bits in one register and the most significant 16 bits in the other register.

In the PLC, since the Comm Format was selected as DATA SINT, which are 8 bit registers, each one of our registers will take up two of the PLC input registers. That is, our word 0 will be split in two parts and be received in the PLC tag locations 0 & 1, our word 1 will be received in tag locations 2 & 3, etc. So each of the above mapped values, since they take two registers each, will take up four tag registers each in the PLC.

Within the program of your PLC, doing a copy command will gather data from the source location into a destination location. (The length will determine how many destinations are filled and the destination tag will determine how much data is gathered from the source location.) This command will not change the data, but rather will literally copy the bit pattern of the data into the destination location. Also realize that the command will gather enough data from the source location to be able to fill the destination location. This means that if you copy from your input table, which is SINT registers, into a DINT (double integer) register, with a length of 1, the PLC will need to take data from four registers at the source to fill the DINT destination register. The same conditions would apply if the destination was a FLOAT register. This means that when the copy command is done with a source of your input table, registers 0 and the destination a DINT registers, then the PLC will take the first four SINT registers' data and place these 32 bits into the DINT destination. Now you have a 32 bit integer value. This same thing happens when you copy from source input register 4 into a FLOAT destination. The PLC will take 32 bits of data starting at SINT register #4 (will take 4-7) and fill the 32 bit FLOAT register.

When the source data is split up into multiple registers and then copied back into one register, the order of the bytes will normally be maintained. This means that when the Hardy unit splits the data into two registers, then it gets split again on the PLC end into SINT registers, the copy command into a DINT or FLOAT register will get all the bits in the proper order to represent the original value based on the data ty