Transaction size in bytes can be roughly calculated using the following formula: (Number of Inputs * 148 bytes) + (Number of outputs * 34 bytes) + 10 bytes
I did not know this before. Very, very useful. Shall use this method already bookmarked to test on my transactions for the next time. I am now just curious: why does the input size look 4 times the site of output? Should not the logic require them to be almost the same size? Or is a lot more data included in input?